Mahesh

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7c

Find the coefficient of

Question 1.
x⁵ in (x + 3)⁸
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = ⁿC_r a^n-rb^r
Here x5 is the T_r+1 term so a = x, b = 3 and n = 8
T_r+1 = 8C_rx^8-r3^r ………..(1)
To find out the coefficient of x⁵, We have to equate x⁵ = x^8-r =>r = 3
Putting the value of r in (1), we get T₃₊₁ = ⁸C₃ × x^8-3 3³
T₄ = \(\frac{8!}{3!5!}\) × x⁵ × 27 = 1512 x⁵
Hence the coefficient of x⁵ = 1512

Question 2.
a⁵b⁷ in (a – 2b)¹².
Solution:
The general term Tr+ t of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-rb^r
Here a = a, b = -2b & n = 12
Substituting above values, we get T_r+1 = ¹²C_r a^12-r (-2b)^r … (1)
To find out the coefficient of a⁵, We equate a^12-r = a⁵, r = 7
Putting r = 7 in (1) ⇒ Tg = ¹²C₇ a⁵(-2b)⁷
T₈ = \(\frac{12!}{7!5!}\) × a⁵ × (-2)⁷b⁷
= -101376 a⁵b⁷
Hence, the coefficient of a⁵b⁷ = -101376.

Write the general term in the expansion of

Question 3.
(x³ – y)⁶
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = nCra”-b’… (1)
Here, a = x², n = 6 and b = – y
Putting above values in (1)
T_r+1 = ⁶C_r x^2(6-r) (-1)^r y^r
= ⁶C_r x^12-2r x (-1)^r X y^r
= (-1)^{r 6}C_r x^12-2r y^r

Question 4.
(x² – yx)¹², x ≠ 0.
Solution:
The general term T_r+x of the binomial expansion (a + b)n is given by T_r+1 = nC_r a^n-r b^r
Here n = 12, a = x² and b = -y x
Substituting the above values, we get
T_r+1 = ¹²C_r x^2(12-r) (-1)^r y^rx^r = (-1)r ¹²C_r x^24-2r y^r x^r
= (-1)^{r 12}C_r x^24-r. y^r

Question 5.
In the expansion of (1 + a)^m+n, prove that the coefficients of am and an are equal.
Solution:
We know that the general term T_r+1 of the binomial expansion (a + b)ⁿ is given by
T_r+1 = ⁿC_r a^n-r b^r
Here n = m + n, a = 1 and b = a
Substituting the values in the general term
T_r+1 = ^m+nC. 1^m+n-ra^r = ^m+nC_r a^r ………..(1)
Now, we have the general term for the expression is, T_r+1 = ^m+nC_r a^r
Now, for coefficient of a^m, T_m+1 = ^m+nC_m a^m
Hence, for the coefficient of m, the value of r = m
So, the coefficient is ^m+nC_m
Similarly, the coefficient of an is ^m+nC_n
^m+nC_m = \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\) And also, ^m+nC_n = \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\)
The coefficient of am and an are same that is \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\)

Question 6.
The coefficients of the (r – 2)^th, r^th and (r + 2)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1:3:5. Find n and r.
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = ⁿC_r a^n-r b^r
Here, the binomial is (1 + x)ⁿ where a = 1, b = x and n = n

The (r + 1)^th term is given by
T_r+1 = ⁿC_r 1^n-r x_r ⇒ T_{(r + 1)} = ⁿC_r x^r
The coefficient of (r + 1)^th term is ⁿC_r

The rth term is given by (r – 1)^th term
T_{(r – 1 – 1)} = ⁿC_r-1
⇒ T_r = ⁿC_{r – 1}x^r-1
∴ The coefficeint of (r – 1)^th term is ⁿC_{r – 1}

For (r – 1)^th term, we will take (r – 2)^th term
T_{r – 2 + 1} = ⁿC_r-2 x^r-2
T_r-1 = ⁿC_r-2x^r-2
∴ The coefficeint of (r – 1)^th term is ⁿC_{r – 2}

Given that the coefficient of (r – 1)^th, r^th and r + 1^th term are in ratio 1: 3 : 5
Therefore,
\(\frac{\text { Coefficient of }(\mathrm{r}-1)^{\text {th }} \text { term }}{\text { Coefficient of } \mathrm{r}^{\text {th }} \text { term }}=\frac{1}{3}\)

On rearrangement, we get
\(\frac{n!}{(r-2)!(n-r+2)!} \times \frac{(r-1)!(n-r+1)!}{n!}=\frac{1}{3}\)

By multiplying
⇒ \(\frac{(\mathrm{r}-1)(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2)!}=\frac{1}{3} \Rightarrow \frac{(\mathrm{r}-1)(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{n}-\mathrm{r}+2)(\mathrm{n}-\mathrm{r}+1)!}=\frac{1}{3}\)

On simplifying we get,
⇒ \(\frac{(\mathrm{r}-1)}{(\mathrm{n}-\mathrm{r}+2)}=\frac{1}{3}\)
⇒ 3r – 3 = n – r + 2
⇒ n – 4r + 5 = 0 …. (1)

Also, \(\frac{\text { the coefficient of } r^{\text {th }} \text { term }}{\text { coefficient of } r+1^{\text {th }} \text { term }}=\frac{3}{5} \Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{3}{5}\)

On rearranging we get,
⇒ \(\frac{n!}{(r-1)!(n-r+1)!} \times \frac{r!(n-r)!}{n!}=\frac{3}{5}\)

By multiplying
⇒ \(\frac{r(r-1)!(n-r)!}{r-1)!(n-r+1)!}=\frac{3}{5} \Rightarrow \frac{r(n-r)!}{(n-r+1)!}=\frac{3}{5} \Rightarrow \frac{r(n-r)!}{(n-r+1)(n-r)!}=\frac{3}{5}\)

On simplifying we get,
⇒ \(\frac{r}{(n-r+1)}=\frac{3}{5}\)
⇒ 5r = 3n – 3r + 3
⇒ 8r – 3n – 3 = 0 … (2)
We have 1 and 2 as
n – 4r + 5 = 0 …(1)
8r – 3n – 3 = 0 …(2)
Multiplying equation (1) by number (2)
2n – 8r + 10 = 0 …(3)
Adding equations (2) and (3)
2n – 8r + 10 = 0
⇒ -3n – 8r – 3 = 0
⇒ -n = -7
⇒ n = 7 and r = 3

Question 7.
Prove that the coefficient of xn in the expansion of (1 + x)²ⁿ is twice the of xⁿ in the expansion of (1 + x)^2n-1.
Solution:
The general term T_r+1, of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r+1 a^n-rb^r
The general term for binomial (1 + x)²ⁿ is
T_(r+1) = ²ⁿC_rx^r …(1)
To find the coefficient of xⁿ, r = n
T_n+1 = ²ⁿC_nxⁿ
The coefficeint of xⁿ = ²ⁿC_n

The general term for binomial (1 + x)^2n-1 is
T_r+1 = ^2n-1C_r x^r

To find the co-efficient of xⁿ
Putting r = n
T_r+1 = ^2n-1C_r x^r
The coefficient of xⁿ = ^2n-1C_n
We have to prove
Coefficient of xⁿ in (1 + x)²ⁿ = 2 coefficient of xⁿ in (1 + x)^2n-1
Consider LHS = ²ⁿC_n = \(\frac{2 n!}{n!(2 n-n)!}=\frac{2 n!}{n!(n)!}\)
Again consider RHS = 2 × ^2n-1C_n
= 2 × \(\frac{(2 n-1)!}{n!(2 n-1-n)!}=2 \times \frac{(2 n-1)!}{n!(n-1)!}\)
Now multiplying and dividing by n we get, 2 × \(\frac{(2 n-1)!}{n!(n-1)!} \times \frac{n}{n}=\frac{2 n(2 n-1)!}{n!n(n-1)!}=\frac{2 n!}{n!n!}\)
From above equation LHS = RHS.
Hence proved.

Question 8.
Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.
Solution:
The general term T_r+1 of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r
Here, a = 1, b = x and n = m
Putting above values, T_r+1 = ^mC_r 1^n-r x^r = ^mC_r x^r
We need the coefficient of x²
Putting r = 2, T₂₊₁ = ^mC₂ x²
The coefficient of x² = ^mC₂
Given that coefficient of x² = ^mC₂ = 6
⇒ \(\frac{\mathrm{m}!}{2!(\mathrm{m}-2)!}\) = 6
⇒ \(\frac{\mathrm{m}(\mathrm{~m}-1)(\mathrm{m}-2)!}{2 \times 1 \times(\mathrm{m}-2)!}\) = 6
⇒ m(m – 1) = 12
⇒ m² – m – 12 = 0
⇒ m² – 4m + 3m – 12 = 0
⇒ m(m – 4) + 3(m – 4) = 0
⇒ (m + 3) (m – 4) = 0
⇒ m = -3, 4
We need the positive value of m, so m = 4.

Practice AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Binomial Theorem MCQ

Question 1.
The 3rd term of (3x – \(\frac{y^3}{6}\))⁴
1) \({ }^4 C_2(3 x)^2\left(\frac{-y^3}{6}\right)^2\)
2) \({ }^4 C_2(3 x)^2\left(\frac{-y^2}{6}\right)^2\)
3) \({ }^4 C_2(3 x)^3\left(\frac{-y^3}{6}\right)^2\)
4) \({ }^4 C_2(3 x)^3\left(\frac{-y^2}{6}\right)^2\)
Answer:
1) \({ }^4 C_2(3 x)^2\left(\frac{-y^3}{6}\right)^2\)

Explanation:
Given (3x – \(\frac{y^3}{6}\))⁴, Here a = 3x, b = \(\frac{-y^3}{6}\), n = 4, T_r+1 = ⁿC_r a^n-r.b^r

T₃ = T₂₊₁ = ⁴C₂ (3x)^4-2\(\left(\frac{-y^3}{6}\right)^2\)
= ⁴C₂ (3x)²\(

Question 2.
C₀ + 3.C₁ + 3².C₂ + …………. + 3ⁿ.C_n =
1) 2ⁿ
2) 3ⁿ
3) 4ⁿ
4) 5ⁿ
Answer:
3) 4ⁿ

Explanation:
We know that (1 + x)ⁿ = C₀ + C₁x + C₁x² + C₃x³ + … + C_nxⁿ … (1)
Put x = 3 in eq (1) we get C₀ + C₁3 + C₂3² + … + C_n3ⁿ = (1 + 3)ⁿ
C₀ + 3.C₁ + 3².C₂ + …… + 3ⁿ.C_n = 4ⁿ

Question 3.
The middle term in the expansion of (4 + 2x)⁶ is
1) 11240 x²
2) 10240 x³
3) 12240 x⁴
4) 10340 x⁴
Answer:
2) 10240 x³

Explanation:
Given (4 + 2x)⁶ contains 7 terms
The middle term T₄ = T₃₊₁ = ⁶C₃ (4)^6-3. (2x)³ = 20 × 64 × 8 × x³ = 10240 x³

Question 4.
The coefficient of x²y² in the expansion of (x + 1)² (y + 1)³ is
1) 3
2) 5
3) 2
4) 10
Answer:
1) 3

Explanation:
(x + 1)² (y + 1)³ = (x² + 2x + 1) (y³ + 3y² + 3y + 1)
Coefficient of x²y² is “3”

Question 5.
In the expansion of (1 + x)²⁰, if the coefficients of rth and (r + 4)^th terms are equal, then r is
1) 7
2) 8
3) 9
4) 10
Answer:
3) 9

Explanation:
Given expansion is (1 + x)²⁰
Given condition is coefficient of r* term = coeff of (r + 4)^th term
⇒ ²⁰C_r-1 = ²⁰C_r+3
⇒ r – 1 + r + 3 = 20 [∵ ⁿC_r = ⁿC_s ⇒ r + s = n]
⇒ 2r = 18
⇒ r = 9

Question 6.
The remainder when 848 is divided by 63 is
1) 4
2) 2
3) 1
4) 7
Answer:
3) 1

Explanation:
848 = (64)²⁴ = (63 + 1)²⁴ = 63q + 1 [∵ q is some Integer]
∴ Remainder = 1.

Question 7.
The expansion of (x + y)¹⁰⁰⁰ is
1) [latex]\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}-1000} \mathrm{y}^{\mathrm{r}}\)
2) \(\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{1000-\mathrm{r}^2} \mathrm{y}^{\mathrm{r}}\)
3) \(\sum_{r=0}^{999}{ }^{1000} C_r x^{r-1000} y^r\)
4) \(\sum_{r=0}^{999}{ }^{1000} C_r x^{1000-r} y^r\)
Answer:
2) \(\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{1000-\mathrm{r}^2} \mathrm{y}^{\mathrm{r}}\)

Explanation:
Given expansion = (x + y)¹⁰⁰⁰ = \(\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{1000-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}\)

Question 8.
The coefficients of the first and the last terms of the expansion (x + y)¹⁰⁰⁰ are
1) 2
2) 3
3) 1
4) 7
Answer:
2) 3

Explanation:
Given expansion is (x + y)ⁿ
(x + y)ⁿ = ⁿC₀xⁿ + ⁿC₁x^n-1y + … + ⁿC_n. yⁿ
Coefficient of first and last terms are ⁿC₀ and ⁿC_n
ⁿC₀ = ⁿC_n = 1.

Question 9.
The expansion of (xy + 2)² is
1) x² + y² + 4
2) x² y² + 4xy + 4
3) xy² + 4 + 2xy
4) x²y² + 2xy + 4
Answer:
2) x² y² + 4xy + 4

Explanation:
(xy + 2)² = (xy)² + 2(xy) . 2 + 2² = x²y² + 4xy + 4

Question 10.
The term ⁹¹C₂x⁸⁹ belongs to
1) x⁸⁹
2) (x – 2)⁹⁰
3) (x – 1)⁹¹
4) (x + 1)⁹⁰
Answer:
3) (x – 1)⁹¹

Explanation:
Term ⁹¹C₂x⁸⁹ is belongs to (x – 1)⁹¹
T₃ of (x – 1)⁹¹ = ⁹¹C₂x^91-2.(-1)²
= ⁹¹C₂x⁸⁹

Question 11.
The cofficient of x⁸y¹⁰ in the expansion f (x + y)¹⁸ is
1) ¹⁸C₈
2) ¹⁸P₁₀
3) 2¹⁸
4) None of these
Answer:
1) ¹⁸C₈

Explanation:
Given expansion is (x + y)¹⁸
The coefficient of x¹⁸y¹⁸ = ¹⁸C₁₀
= ¹⁸C_18-10
= ¹⁸C₈

Practice AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Permutations and Combinations MCQ

Question 1.
A student has 5 pants and 8 shirts. The number of ways in which he can wear the dress in different combinations as
1) ⁸P₅
2) ⁸C₅
3) 8! × 5!
4) 40
Answer:
4) 40

Explanation:
Student can select a pant from 5 pants in 5 ways and a shirt from 8 shirts in 8 ways.
∴ The number of ways he can wear the dress in different combinations = 5 × 8 = 40.

Question 2.
If ⁿP₄ = 1680 then n is
1) 7
2) 8
3) 9
4) 10
Answer:
2) 8

Explanation:
ⁿP₄ = 1680 = 8 × 7 × 6 × 5
∴ n = 8

Question 3.
If ⁿP_r = 1320, then r is
1) 3
2) 4
3) 5
4) 6
Answer:
1) 3

Explanation:
ⁿP_r = 1320 = 12 × 11 × 10
∴ r = 3

Question 4.
A man has 3 sons and 6 schools within his reach. In how many ways he can send his sons to school if no two of his sons are to read in the same school ?
1) ⁶P₃
2) ⁶P₃
3) 6³
4) 3⁶
Answer:
1) ⁶P₃

Explanation:
The number of ways that he can send his 3 sons to 6 schools if no two of then are in the same school = ⁶P₃

Question 5.
The number of arrangements that can be formed by talking all the letters {T, U, E, S, D. A, Y}
1) 720
2) 5040
3) 120
4) 40320
Answer:
2) 5040

Explanation:
The number of required arrangements = 7! = 5040

Question 6.
The number of arrangements that can be made out of the letters in expresesion a⁴b³c⁵ when written in full length is
1) \(\frac{9!}{3!2!4!}\)
2) \(\frac{7!}{2!3!2!}\)
3) \(\frac{5!}{4!3!2!}\)
4) \(\frac{(12)!}{4!3!5!}\)
Answer:
4) \(\frac{(12)!}{4!3!5!}\)

Explanation:
The required number of arrangements = \(\frac{(12)!}{4!3!5!}\)

Question 7.
The number of permutations of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 when repitition allowed is
1) ⁶P₄
2) ⁶C₄
3) 6⁴
4) 4⁶
Answer:
3) 6⁴

Explanation:
The number fo 4-digit numbers formed = 6⁴.

Question 8.
The number of ways of arranging 8 men and 4 women in a row is
1) 8!
2) 4!
3) (12)!
4) (11)!
Answer:
3) (12)!

Explanation:
The number of ways of arranging 8 men and 4 women in a row = 12!

Question 9.
The number of positive divisors of 1080 is
1) 8
2) 16
3) 32
4) 40
Answer:
3) 32

Explanation:
1080 = 2³ × 3³ × 5¹ ⇒ The number of positive divisors of 1080 is
(3 + 1) (3 + 1) (1 + 1) = 32.

Question 10.
The number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls is
1) ⁸C₄ + ⁵C₃
2) ⁸C₄ × ⁵C₃
3) ⁸C₃ – ⁵C₃
4) None
Answer:
2) ⁸C₄ × ⁵C₃

Explanation:
The number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls is ⁸C₄ × ⁵C₃

Question 11.
If ¹²C_s+1 = ¹²C_2s-5 then the value of S is
1) 4
2) 8
3) 12
4) 6
Answer:
4) 6

Explanation:
¹²C_s+1 = ¹²C_2s-5 ⇒ S + 1 = 2S – 5 or S + 1 + 2S – 5 = 12
S = 6 or 3S = 16
S = 6 [∵ S is a non-negative integer]

Question 12.
The number of diagonals of a polygon of 10 sides is
1) 35
2) 36
3) 38
4) 39
Answer:
1) 35

Explanation:
Number of diagonals of a polygon of 10 sides = \(\frac{10(10-3)}{2}\) = 35

Question 13.
A polygon has 54 diagonals. Then the number of its sides is
1) 7
2) 9
3) 10
4) 12
Answer:
4) 12

Explanation:
If the number of sides is 4 then the number of diagonals
\(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 54
⇒ n = 12.

Referring to the AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations Exercise 6e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6e

I.

Question 1.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of i) exactly 3 girls ii) atleast 3 girls iii) atmost 3 girls.
Solution:
A committee of 7 has to be formed from 9 boys and 4 girls.
i) Since exactly 3 girls are to be there in every committee. Each committee must consist of (7 – 3) = 4 boys only.
Thus, in this case, required number of ways
= ⁴C₃ × ⁹C₄
= \(\frac{4!}{3!1!} \times \frac{9!}{4!5!}\)
= 4 × \(\frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!}\)
= 508

ii) Since atleast 3 girls are to be there in every committee. The committee can consist of
a) 3 girls and 4 boys
b) 4 girls and 3 boys
3 girls and 4 boys can be selected in ⁴C₃ × ⁹C₄ ways.
4 girls and 3 boys can be selected in ⁴C₂ × ⁹C₅ ways.
Therefore, in this case, required number of ways.
= ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃
= 504 + 84
= 588
Therefore, in this case, required number of ways.
= ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃
= 504 + 84
= 588

iii) Since atmost 3 girls are to be there in every committee. The committee can consist of
a) 3 girls and 4 boys
b) 2 girls and 5 boys
c) 1 girl and 6 boys
d) No girl and 7 boys
3 girls and 4 boys can be selected in ⁴C₃ × ⁹C₄ ways.
2 girls and 5 boys can be selected in ⁴C₂ × ⁹C₅
1 girl and 6 boys can be selected in ⁴C₁ × ⁹C₆
No girl and 7 boys can be selected in ⁴C₀ × ⁹C₇ ways.
Therefore, in this case, required number of ways.
= ⁴C₃ × ⁹C₄ + ⁴C₂ × ⁹C₅ + ⁴C₁ × ⁹C₆ + ⁴C₀ × ⁹C₇
= \(\frac{4!}{3!1!} \times \frac{9!}{4!5!}+\frac{4!}{2!2!} \times \frac{9!}{5!4!}+\frac{4!}{1!3!} \times \frac{9!}{6!3!}+\frac{4!}{0!4!} \times \frac{9!}{7!2!}\)
= 504 + 756 + 336 + 36
= 1632

Question 2.
It is required to seat 5 .men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?
Solution:
5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways = M × M × M × M × M.
Therefore, the women can be seated in 4! ways.
Hence, possible number of arrangements = 4! × 5! = 24 × 120 = 2880.

Question 3.
In how many ways can be letters of the word ASSASSINATION be arranged so that all the S’s are together ?
Solution:
In the given word ASSASSINATION, the letter A appear 3 times.
S appears 4 times, I appear 2 times, N appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the ‘S’ are together, SSSS is treated as a single objects for the time being. This single object together with the remaining 9 objects will account for 10 objects.
The number of arrangements of 10 objects in which there are 3’A’, 2T and 2’N’.
\(\frac{10!}{3!2!2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3!\times 2 \times 1 \times 2 \times 1}\) = 151200.
Hence, the required number of ways of arranging the letters of the given word is 151200.

II.

Question 1.
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Solution:
In the word DAUGHTER, there are 3 vowels namely A, U and E and 5 consonants, namely D, G, H, T and R.
Number of ways of selecting 2 vowels out of 3 vowels = ³C₂ = 3.
Number of ways of selecting 3 consonants out of 5 consonants = ⁵C₃ =10. Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30.
Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.
Hence, required number of different words = 30 × 5! = 3600.

Question 2.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together ?
Solution:
In the word EQUATION, there are 5 vowels, namely, A, E, I, O and U and 3 consonants namely Q, T and N.
Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted.
This number would be ²P₂ = 2!
Corresponding to each of these permutations, there are 5! Permutations of the 5 vowels taken all at a time and 3! Permutations of the 3 consonants taken all at a time.
Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440.

Question 3.
If the different permutations of all the letters of the word EXAMINATION are listed in a dictionary, how many words are there in this list before first word start with E ?
Solution:
In the given word EXAMINATION there are 11 letters out of which A, I and N appear 2 times and all the other letters appears only once.
The words that will be listed before the word starting with E in a dictionary will be word that start with A. Therefore to get the number of words starting with E in a dictionary will be the words that start with A since there are 2T and 2 ’N’ in the remaining 10 letters.
Number of words starting with A = \(\frac{10!}{2!2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2 \times 1 \times 2!}\)
= 907200
Hence, the required number of words are 907200.

Question 4.
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?
Solution:
A number is divisible by 10 if its unit’s digits is 0.
Therefore, 0 is fixed at the unit’s place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places

In succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9)
The 5 vacant places can be filled in 5! ways.
Hence, required number of 6-digits = 5! = 120.

Question 5.
The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Solution:
2 different vowels and 2 different consonants are to be selected from the English alphabet.
Since there are 5 vowels in the English alphabet.
Number of ways of selecting 2 different vowels = ⁵C₂ = \(\frac{5!}{2!3!}=\frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}\) = 10
Since there are 21 consonants in the English alphabet.
Number of ways of selecting 2 different consonants from the alphabet
= ²¹C₂ = \(\frac{21!}{2!9!}=\frac{21 \times 20 \times 19!}{2 \times 1 \times 19!}\) = 210
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 x 210 = 2100
Each of these 2100 combinations has 4 letters which can be arranged among themselves in 4! ways.
Therefore, required number of words = 2100 × 4! = 50400.

III.

Question 1.
In an examination, a question paper consists of 12 questions divided into two parts i.e. part I and part II containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Solution:
It is given that the question paper consists of 12 questions divided into two parts – part I and and part II containing 5 and 7 questions, respectively.
A student has to attempts 8 questions at least 3 from each part. This can be done as follows.
a) 3 questions from part I and 5 questions from part II
b) 4 questions from part I and 4 questions from part II
c) 5 questions from part I and 3 questions from part II
3 questios from part I and 5 questions from part II can be selected in ⁵C₃ × ⁷C₄ ways.
4 questions from part I and 4 questions from part II can be selected in ⁵C₄ × ⁷C₄ ways.
5 questions from part I and 3 questions from part II can be selected in ⁵C₅ × ⁷C₂ ways
= ⁵C₃ × ⁷C₄ + ⁵C₄ × ⁷C₄ + ⁵C₅ × ⁷C₂

Question 3.
From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?
Solution:
From the class of 25 students, 10 are to be chosen for an excursion party.
Since there are 3 students who decide that either all of them will join or none of them will join, there are two cases.
Case-I : All of three students join.
Then, 10 students can be chosen from the remaining 22 students ²²C₇ ways.
Case-II : None of the three students join.
Then, 10 students can be chosen from the remaining 22 students in ²²C₁₀ ways. Hence, the required number of ways of choosing the excursion party is ²²C₇ + ²²C₁₀

Referring to the AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations Exercise 6d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6d

Question 1.
If ⁿC₈ = ⁿC₂ find ⁿC₂.
Solution:
We Know that if ⁿC_a = ⁿC_b, then a = b or a + b = n
Therefore, ⁿC₈ = ⁿC₂
⇒ 8 + 2 = n
⇒ n = 10
Hence, ⁿC₂ = ¹⁰C₂
= \(\frac{10!}{2!(10-2)!}=\frac{10!}{2!8!}=\frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}\)
= 5 × 9
= 45

Question 2.
Dterminw n if i) ²ⁿC₃ : ⁿC₃ = 12 : 1
Solution:
Given that ²ⁿC₃ : ⁿC₃ = 12 : 1
⇒ \(\frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{12}{1}\)

⇒ 8n – 4 = 12n – 24
⇒ 4n = 20
⇒ n = 5

ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Solution:
²ⁿC₃ : ⁿC₃ = 11 : 1

⇒ 8n – 4 = 11n – 22
⇒ 3n = 18
⇒ n = 6

Question 3.
How many chords can be drawn though 21 points on a circle ?
Solution:
Number of chords that can be drawn though 21 points on a circle
= Number of ways of selecting 2 points from 21 points = ²¹C₂ = \(\frac{21 \times 20}{2}\) = 210.

Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls ?
Solution:
3 boys out of 5 boys can be selected in ⁵C₃ ways.
3 girls out of 4 girls can be selected in ⁴C₃ ways.
∴ No. of ways in which 3 boys and 3 girls are selected = ⁵C₃ × ⁴C₃
= 10 × 4
= 40 ways.

Question 5.
If ⁿC₄ = 210, find n.
Solution:
Given that ⁿC₄ = 210
⇒ \(\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}\) = 10 × 21
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 21 × 1 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 7 × 3 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 9 × 8 × 7
⇒ n = 10

Question 6.
If ¹²C_r = 495, find the possible values of ‘r’.
Solution:
¹²C_r = 495
= 5 × 99
= 11 × 9 × 5
= \(\frac{12 \times 11 \times 9 \times 5 \times 2}{12 \times 2}=\frac{12 \times 11 \times 10 \times 9}{1 \times 2 \times 3 \times 4}\)
= ¹²C₄ or ¹²C₈
∴ r = 4 or 8

Question 7.
If 10.ⁿC₄ = 3. ⁿ⁺¹C₃, find n.
Solution:
Given that 10.ⁿC₂ = 3.ⁿ⁺¹C₃
⇒ 10 × n(n – 1)1.2 = 3(n + 1)(n – 1)1.2.3
⇒ 10 = n + 1
⇒ n = 9

Question 8.
If ¹⁵C_2r-1 = ¹⁵C_2r+4 + 4, find r.
Solution:
Given that ¹⁵C_2r-1 = ¹⁵C_2r+4
⇒ 2r – 1 = 2r + 4 or (2r – 1) + (2r + 4) = 15
⇒ (2r – 1) + (2r + 4) = 15
⇒ 4r + 3 = 15
⇒ 4r = 12
⇒ r = 3
∴ r = 3

Question 9.
If ¹²C_r+1 = ¹²C_3r-5, find r.
Solution:
Given that ¹²C_r+1 = ¹²C_3r-5
⇒ r + 1 = 3r – 5 or (r + 1) + (3r – 5) = 12
⇒ 1 + 5 = 2r or 4r – 4 = 12
⇒ 2r = 6 or 4r = 16
⇒ r = 3 or r = 4
r = 3 or 4

Question 10.
If ⁿC₅ = ⁿC₆, then find ¹³C₄.
Solution:
Given that ⁿC₅ = ⁿC₆
⇒ n = 6 + 5 = 11
⇒ ¹³C_n = ¹³C₁₁
= ¹³C₂
= \(\frac{13 \times 12}{1 \times 2}\) = 78

II.

Question 1.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection Solution:
The no. of ways of selecting 3 red balls out of 6 red balls = ⁶C₃
The no. of ways of sleeting 3 white balls out of 5 white balls = ⁵C₃
The no. of ways of selecting 3 blue balls out of 5 blue balls = ⁵C₃
The no. of ways of selecting 3 balls of each colour = ⁶C₃ × ⁵C₃ × ⁵C₃
= 20 × 10 × 10
= 2000.

Question 2.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution:
One ace will be selected from four aces and four cards will be selected from 52 cards, then 52 – 4 = 48 cards.
If ‘p’ is the required no. of ways, then
⇒ P = ⁴C₁ × ⁴⁸C₄
= 4 × \(\frac{48!}{4!(48-4)!}=\frac{4 \times 48!}{4!44!}\)
= 4 × \(\frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 1 \times 44!}\)
= 778320 ways

Question 3.
In how many ways can one select a cricket team of Eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers ?
Solution:
Out of 17 players, 5 players are bowlers. A cricket team of 11 players is to be selected in such away that there are exactly 4 bowlers. 4 bowlers can be selected out of 5 bowlers in 5C4 ways and the remaining 7 players can be selected out of the 12 players in 12C7 ways.
Thus, by multiplication principle, required number of ways of selecting circket team
⁵C₄ × ¹²C₇ = \(\frac{5!}{4!(5-4)!} \times \frac{12!}{7!(12-7)!}=\frac{5 \times 4!}{4!1!} \times \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{7!5!}\)
= 5 × 792
= 3960

Question 4.
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Solution:
There axe 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in ⁵C₂ ways and
3 red balls can be selected out of 6 red balls in ⁶C₃ ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls.
⁵C₂ × ⁶C₄
= \(\frac{5!}{2!(5-2)!} \times \frac{6!}{3!(6-3)!}=\frac{5 \times 4 \times 3!}{2!3!}=\frac{6 \times 5 \times 4 \times 3!}{3!3!}\)
= 10 × 20
= 200.

Question 5.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student ?
Solution:
There are 9 courses available out of which, 2 specific courses are compulsory for every student and has to choose 3 courses out of the remaining 7 courses. 2 courses can be selected out of 2 specific courses in ²C₂ ways and 3 courses can be selected out of the remaing 7 courses in ⁷C₃ ways.
Thus, the required number of ways of choosing the programme
²C₂ × ⁷C₃ = \(\frac{2!}{2!(2-2)!} \times \frac{7!}{3!(7-3)!}=\frac{2!}{2!0!} \times \frac{7 \times 6 \times 5 \times 4!}{3!4!}\)
= 1 × 35
= 35

Question 6.
Prove that for 3 ≤ r ≤ n.
^{(n – 3)}C_r + 3.^{(n – 3)}C_{(r – 1)} + 3. ^{(n – 3)}C_{(r – 2)} + ^{n – 3}C_{r – 3} = ⁿC_r
Solution:
Since ⁿC_{r – 1} + ⁿC_r = ^{n + 1}C_r, Given can be taken as
^{n – 3}C_r + ^{n – 3}Cr + 2[^{n – 3}C_{r – 1} + ^{n – 3}C_{r – 2}] + ⁿ C_{r – 2} + ^{n – 3}C_{r – 3}
= ^{n – 2}C_r + 2.^{(n – 2)}C_{r – 1} + ^{n – 2}C_{r – 2}
= ^{n – 1}C_r + ^{n – 1}C_{r – 1}
= ^{n – 1 + 1}C_r
= ⁿC_r

Question 7.
Simp;ify ³⁴C₅ + \(\sum_{\mathbf{r}=0}^4\) ^{(38 – r)}C₄.
Solution:
³⁴C₅ + \(\sum_{\mathbf{r}=0}^4\) ^{(38 – r)}C₄.

Question 8.
In a class there are 30 students. If each student plays a chess game with each of the other student, then find the total number of chess games played by them.
Solution:
Number of students in a class = 30.
Since each student plays a chess game with each of the other students, the total
number of chess games played by them = ³⁰C₂ = 435

Question 9.
Find the number of ways of selecting 3 vowels and 2 consonants from the letters of the word EQUATION.
Solution:
The word EQUATION contains 5 vowels and 3 consonants.
The 3 vowels can be selected from 5 vowels in ⁵C₃ = 10 ways.
The 2 consonants can be selected from 3 consonants in ³C₂ = 3 ways.
The required number of ways of selecting 3 vowels and 2 consonants = 10 × 3 = 30.

Question 10.
Find the number of diagonals of a polygon with 12 sides.
Solution:
The number of diagonals of a polygon with 12 sides = \(\frac{n(n-3)}{2}=\frac{12(12-3)}{2}\) = 54

Question 11.
If n persons are sitting in a row, find the number of ways of selecting two persons, who are sitting adjacent to each other.
Solution:
The number of ways of selecting 2 persons out of n persons sitting in a row, who are sitting adjacent to each other = n – 1

III.

Question 1.
Prove that \(\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1.3 .5 \ldots(4 n-1)}{\{1.3 .5 \ldots(2 n-1)\}^2}\).
Solution:
im-4

Question 2.
If a set A has 12 elements, find the number of subsets of A having
(i) 4 elements
(ii) Atleast 3 elements
(iii) Atmost 3 elements.
Solution:
Number of elements in set A = 12.
i) Number of subsets of A with exactly 4 elements ¹²C₄ = 495.
ii) The required subset contains atleast 3 elements.
The number of subsets of A with exactly 0 elements is ¹²C₀
The number of subsets of A with exactly 1 element is ¹²C₁
The number of subsets of A with exactly 2 elements is ¹²C₂
Total number of subsets of A formed = 2¹².
Number of subsets of A with atleast 3 elements = (Total number of subsets) – (number of subsets contains 0 or 1 or 2 elements)
= 2¹² – (¹²C₀ + ¹²C₁ + ¹²C₂)
= 4096 – (1 + 12 + 66)
= 4096 – 79
= 4017

iii) The required subset contains atmost 3 elements.
i.e., it may contain 0 or 1 or 2 or 3 elements.
The number of subsets of A with exactly 0 elements is ¹²C₀.
The number of subsets of A with exactly 1 elements is ¹²C₁
The number of subsets of A with exactly 2 elements is ¹²C₂.
The number of subsets of A with exactly 3 elements is ¹²C₃.
∴ Number of subsets of A with atmost 3 elements = ¹²C₀ + ¹²C₁ + ¹²C₂ + ¹²C₃
= 1 + 12 + 66 + 220
= 299

Question 3.
There are 8 railway stations along a railway line. In how many ways can a train be stopped at 3 of these stations such that no two of them are consecutive ?
Solution:
Number of ways of selecting 3 stations out of 8 = ⁸C₃ = 56.
Number of ways of selecting 3 out of 7 stations such that 3 are consecutive = 6. Number of ways of selecting 3 out of 8 stations such that 2 of them are consecutive = 2 × 5 + 5 × 4
= 10+ 20
= 30
Number of ways for a train to be stopped at 3 or 8 stations such that no two of them are consecutive = 56 – (6 + 30) = 20.

Question 4.
A question paper is divided into 3 sections A, B, C containing 3, 4, 5 questions respectively. Find the number of ways of attempting 6 questions choosing atleast one from each section.
Solution:
The selection of a question may be of the following

Total No. of ways of attempting 6 questions
= ³C₃. ⁴C₂.⁵C₁ + ³C₃. ⁴C₁.⁵C₂ + ³C₂. ⁴C₃.⁵C₁ + ³C₂. ⁴C₂.⁵C₂ + ³C₂. ⁴C₁.⁵C₃ + ³C₁. ⁴C₄.⁵C₁ + ³C₁. ⁴C₃.⁵C₂ + ³C₁. ⁴C₂.⁵C₃ + ³C₁. ⁴C₁.⁵C₄
= 805

Question 5.
A class contains 4 boys and g girls. Every Sunday, five students with atleast 3 boys go for a picnic. A different group is being sent every week. During the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed is 85, find g.
Solution:
No. of boys = 4, No. of girls = g
Since there should be atleast 3 boys it can be done in 2 ways as shown in the table.
The number of girls in G₁ = [⁴C₃ × ^gC₂] × 2
Since each group contains 2 girls.
The number of girls in G₂ = [⁴C₄ × ^gC₁] × 1.
Since each group contains 1 girl.

Given no. of dolls distributed = 85
⇒ [⁴C₃ × ^gC₂] × 2 + [⁴C₄ × ^gC₁] × 1
⇒ 4. g(g – 1)2 × 2 + 1.g.1 = 85
⇒ 4g(g – 1) + 17(g – 5) = 0
⇒ (g – 5)(4g + 17) = 0
Since g ≠ –\(\frac{17}{4}\)
∴ g = 5 ⇒ Hence No. of girls = 5

Referring to the AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations Exercise 6c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6c

I.

Question 1.
If ⁿP₃ = 1320, find n.
Solution:
Given that ⁿP₃ = 1320
\(\frac{n!}{(n-3)!}\) = 1320

= 132 × 10
⇒ n(n – 1)(n – 2)
= 12 × 11 × 10
Comparing on both sides, we get
∴ n = 12

Question 2.
If ⁿP₇ = 42, find n.
Solution:
Given that ⁿP₇ = 42. ⁿP₅

⇒ (n – 5)(n – 6) = 7 × 6
⇒ n – 5 = 7 or n – 6 = 6

Question 3.
If ⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7, find n.
Solution:
Given that ⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7
⇒ \(\frac{(n+1)!}{(n-4)!} \times \frac{(n-6)!}{n!}=\frac{2}{7}\)

⇒ \(\frac{n+1}{(n-4)(n-5)}=\frac{2}{7}\)
⇒ 7n + 7 = 2(n² – 9n + 20)
⇒ 2n² – 25n + 33 = 0
⇒ (n – 1) (2n – 3) = 0
⇒ n – 11 = 0 or 2n – 3 = 0
∴ n = 11

Question 4.
If ¹²P₅ + 5. ¹²P₄ = ¹³P_r, find r.
Solution:
Given that ¹²P₅ + 5. ¹²P₄ = ¹³P_r
⇒ ¹³P₅ = ¹³P_r [∵ ⁿP_r + r. ⁿP_r-1 = ⁽ⁿ⁺¹⁾P_r]
⇒ r = 5 [∵ ⁿP_r = ⁿP_s ⇒ r = s or r + s = n]

Question 5.
If ¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7, find r.
Solution:
Given that ¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7
⇒ \(\frac{{ }^{18} P_{r-1}}{{ }^{17} P_{r-1}}=\frac{9}{7}\)

⇒ \(\frac{18!}{[18-(r-1)]} \times \frac{[17-(r-1)]!}{17!}=\frac{9}{7}\)

⇒ \(\frac{18 \times 17!}{(19-r)!} \times \frac{(18-r)!}{17!}=\frac{9}{7}\)

⇒ \(\frac{18(18-r)!}{(19-r)!}=\frac{9}{7}\)

⇒ \(\frac{18(18-r)!}{(19-r)(18-r)!}=\frac{9}{7}\)

⇒ \(\frac{18}{19-r}=\frac{9}{7}\)
⇒ 19 – r = 14
⇒ r = 19 – 14
⇒ r = 5

Question 6.
Find ‘n’, If ^n-1P₃ : ⁿP₄ = 1 : 9
Solution:
Given that \(\frac{{ }^{n-1} P_3}{{ }^n P_4}=\frac{1}{9}\)

⇒ \(\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}\)

⇒ \(\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!}=\frac{1}{9}\)

⇒ \(\frac{(n-1)!}{n \times(n-1)!}=\frac{1}{9}\)

⇒ \(\frac{1}{n}=\frac{1}{9}\)
⇒ n = 9

Question 7.
Find r, if
i) ⁵P_r = ²P_r-1 = 2 ⁶P_r-1
Solution:
\(\frac{5!}{(5-r)!}=\frac{2 \times 6!}{(6-r+1)!}\)

⇒ \(\frac{5!}{(5-r)!}=\frac{2 \times 6 \times 5!}{(7-r)!}\)

⇒ \(\frac{5!}{(5-r)!}=\frac{2 \times 6 \times 5!}{(7-r)(6-r)(5-r)!}\)

⇒ 1 = \(\frac{2 \times 6}{(7-r)(6-r)}\)
⇒ (7 – r)(6 – r) = 12
⇒ 42 – 6r – 7r + r² = 12
⇒ r² – 13r + 30 = 0
⇒ r² – 3r – 10r + 30 = 0
⇒ r(r – 3) – 10(r – 3) = 9
⇒ (r – 3)(r – 10) = 0
ⁿP_r = \(\frac{n!}{(n-r)!}\) where 0 ≤ r ≤ n
0 ≤ r ≤ 5
∴ r = 3 but r ≠ 10

ii) ⁵P_r = ⁶P_r-1
Solution:
⇒ \(\frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}\)

⇒ \(\frac{5!}{(5-r)!}=\frac{6 \times 5!}{(7-r)!}\)

⇒ \(\frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!}\)

⇒ 1 = \(\frac{6}{(7-r)(6-r)}\)
⇒ (7 – r)(6 – r) = 6
⇒ 42 – 7r – 6r + r² – 6 = 0
⇒ r² – 13r + 36 = 0
⇒ r² – 4r – 9r + 36 = 0
⇒ r(r – 4) – 9(r – 4) = 0
⇒ (r – 4)(r – 9) = 0
⇒ r – 4 = 0 or r – 9 = 0
When ⁿP_r = \(\frac{n!}{(n-r)!}\) where 0 ≤ r ≤ n,
∴ 0 ≤ r ≤ 5
Hence, r ≠ 9, ∴ r = 4

Question 8.
A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons in the schools so that no two of them will be in the same school.
Solution:
The number of ways of admitting 4 sons into 5 schools if no two of them will be in the same school = ⁵P₄ = 5 × 4 × 3 × 2 = 120

Question 9.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated ?
Solution:
3 digit numbers have to be formed using the digits 1 to 9.
Here, the order of the digit’s matters.
Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.
So, the required number of 3-digit numbers 9!
= ⁹P₃ = \(\frac{9!}{(9-3)!}=\frac{9!}{6!}=\frac{9 \times 8 \times 7 \times 6!}{6!}\) = 9 × 8 × 7 = 504.

Question 10.
How many 4-digit numbers are there with no-digit repeated ?
Solution:
The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 6 cannot be included. Therefore, the number of ways in which thousands place can be filled with 1 to 9.
The hundreds, tens and unit places can be filled with any of the digits from 0 to 9.
However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit. The hundred’s, ten’s and unit places are to be filled by the remaining 9 digits.
Therefore, there will be as many such different ways as there are permutations of 9 different digits taken 3 at a time.
So, number of ways to fill remaining 3 places
= ⁹P₃ = \(\frac{9!}{(9-3)!}=\frac{9!}{6!}=\frac{9 \times 8 \times 7 \times 6!}{6!}\)

= 9 × 8 × 7 = 504.
Hence by multiplication principle, the required number of 4-digit numbers is 9 × 504 = 4536.

Question 11.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digits is repeated ?
Solution:
3-digit even numbers are to be formed using the given six digits 1, 2, 3, 4, 6 and 7, without repeating the digits.

Then, unit’s digits can be filled in 3 ways with any of the digits 2, 4 or 6.
Since, the digits cannot be repeated in the 3-digit numbers and units place is already occupied with 9.
Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.
So, number of ways of filling hundreds and tens place.
= ⁵P₂ = \(\frac{5!}{(5-2)!}=\frac{5!}{3!}=\frac{5 \times 4 \times 3!}{3!}\)
= 5 × 4 = 20
Hence, by multiplication principle, the required number of 3 digit even numbers = 3 × 20 = 60.

Question 12.
Find the number of 4 digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even ?
Solution:
4 digit numbers are to be formed using the digits 1, 2, 3, 4 and 5.
There will be as many 4 digit number as there are permutations of 5 different digits taken 4 at a time.
Therefore, required numbers of 4 digit numbers
⁵P₄ = \(\frac{5!}{(5-4)!}=\frac{5!}{1!}=\frac{5 \times 4 \times 3 \times 2 \times 1}{1}\) = 120
Among the 4-digit numbers formed by using the digits 1, 2, 3, 4, 5, even numbers end with either 2 or 4.
The number of ways in which units place is filled with digits is 2.
Since the digits are not repeated and the units place is already occupied with a
digit (which is even) the remaining places are to be filled by the remaining 4 digits.
Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.
Number of ways of filling the remaining places = ⁴P₃ = \(\frac{4!}{(4-3)!}\) = 4 × 3 × 2 × 1 = 24
Thus by multiplication principle, the required number of even numbers is 24 × 2 = 48.

Question 13.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice-chairman assuming one person cannot hold more than one position?
Solution:
From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Here, the number of ways choosing a chairman and vice-chairman is the permutation of 8 different objects taken 2 at a time.
Therefore, required number of ways = ⁸P₂ = \(\frac{8!}{(8-2)!}=\frac{8!}{6!}=\frac{8 \times 7 \times 6!}{6!}\) = 56

Question 14.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed.
Solution:
The number of 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed = 6⁴ = 1296.

Question 15.
Find the number of 5 letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times.
Solution:
The number of 5 letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times = 5⁵ = 3125.

Question 16.
Find the number of functions from a set A containing 5 elements into a set B containing 4 elements. (Hint : The total number of functions from set A containing m elements to set B containing n elements is n^m.)
Solution:
Set A contains 5 elements. Set B contains 4 elements.
For the image of each of the 5 elements of the set, A has 4 choices.
∴ The number of functions from set A containing 5 elements into a set B containing 4 elements = 4 × 4 × 4 × 4 × 4 (5 times) = 45 = 1024.

Question 17.
Find the number of ways of arranging the letters of the words,
(i) MATHEMATICS
(ii) SINGING
(iii) PERMUTATION
(iv) COMBINATION
(v) INTERMEDIATE
Solution:
(i) MATHEMATICS : The word MATHEMATICS contains 11 letters in which there are 2 M’s are alike, 2 A’s are alike, 2 T’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{11!}{2!2!2!}\)

(ii) SINGING : The word SINGING contains 7 letters in which there are 2 I’s are alike, 2. N’s are alike, 2 G’s are alike and rest is different.
∴ The number of required arrangements = \(\frac{7!}{2!2!2!}\)

(iii) PERMUTATION : The word PERMUTATION contains 11 letters in which there are 2 T’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{11!}{2!}\)

(iv) COMBINATION : The word COMBINATION contains 11 letters in which there are 2 O’s are alike, 2 I’s are alike, 2 N’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{11!}{2!2!2!}\)

(v) INTERMEDIATE : The word INTERMEDIATE contains 12 letters in which there are 2 I’s are alike, 2 Ts are alike, 3 E’s are alike and rest are different.
∴ The number of required arrangements = \(\frac{12!}{2!2!3!}\)

Question 18.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together ?
Solution:
In the given word MISSISSIPPI, I appear 4 times, S appears 4 times, P appears 2 times, and M appears just once.
Therefore, number of distinct permutation of letters in the given word.
\(\frac{11!}{4!4!2!}=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!\times 4 \times 3 \times 2 \times 1 \times 2 \times 1}\)
= 11 × 10 × 9 × 7 × 5
= 34650
There are 4 Ts in the given word when they occur together, they are treated as a single object for the time being. This single object Till’ together with the remaining 7 objects will account 8 objects.
These 8 objects in which there 4 ‘S’ and 2 ‘P’ can be arranged in
\(\frac{8!}{4!2!}=\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!\times 2 \times 1}\)
= 840 ways
Number of arrangements where all ‘I’ occur together = 840. Hence, number of distinct permutations of the letter in MISSISSIPI in which four T do not come together = 34650 – 840 = 33810.

Question 19.
If there are 25 railway stations on a railway line, how many types of single second class tickets must be printed, so as to enable a passenger to travel from one station to another.
Solution:
Number of stations on a railway line 25.
∴ The number of single second class tickets must be printed so as to enable a passenger to travel from one station to another ²⁵P₂ = 25 × 24 = 600

Question 20.
Find the number of ways of arranging the letters to the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.
Solution:
Vowels A, E, I, O, U.
In a given word, the number of vowels is 3, number of consonants is 5

Since the relative positions of the vowels and consonants are not disturbed, the 3 vowels can be arranged in their relative positions in 3! ways and the 5 consonants can be arranged in their relative positions in 5! ways.
∴ The number of required arrangements = (3!) (5!) = 6 × 120 = 720

II.

Question 1.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
i) 4 letters are used at a time
ii) all letters are used at a time
iii) all letters are used but first letter is a vowel ?
Solution:
The word MONDAY contains 6 different letters.
i) Number of 4 letter words that can be formed from the letters of the word MONDAY, without repetition of letters is the number of permutations of 6 different objects taken 4 at a time is
⁶P₄ = \(\frac{6!}{(6-4)!}=\frac{6!}{2!}\) = 6 × 5 × 4 × 3 = 360.

ii) Number of words that can be formed by using all the letters of word MONDAY at a time is the number of permutations of 6 different objects taken 6 at time is ⁶P₆ = 6! = 720.

iii) In the given word, there are 2 different vowels, which have to occupy the right most place of the words formed in 2 ways only. The letters cannot be repeated and right most place is already occupied with a letter, the remaining five places are to be filled by the remaining 5 letters is 5! ways.
∴ The required number of words that can be formed is 5! × 2 = 120 × 2 = 240.

Question 2.
In how many ways can the letters of the word PERMUTATIONS be arranged if the
i) words start with P and end with S,
ii) vowels are all together
iii) there are always 4 letters between P and S ?
Solution:
i) If P and S are fixed at the extreme ends (P at the left end S at the right end), then 10 letters are left. So in this case, required number of arrangements are
\(\frac{10!}{2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!}\) = 1814400

ii) There are 5 vowels in the given word. Each appearing only once since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 2 ‘T’ can be arranged in a \(\) ways.
Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways. Therefore, the required number of arrangements can be
\(\frac{8!}{2!}\) × 5!
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!}\) × 120 = 2419200

iii) The letters have to be arranged in such a way that there are always 4 letters P and S. Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in \(\frac{10!}{2!}\) which there are 2 ‘T’ can be arranged in yyy ways. For the places P and S, there are 7 following cases
P and S can take the position of 1 and 6, P and S can take the position of 2 and 7. P and S can take the position of 3 and 8, P and S can take the position of 4 and 9. P and S can take the position of 5 and 10, P and S can take the position of 6 and 11. P and S can take the position of 7 and 12.
Also the letters P and S can interchange in two ways.
So, the ways to arrange P and S such that there are 4 letters between them in 2 × 7 = 14 ways.
Therefore, by multiplication principle, required number of arrangements in this case is
\(\frac{10!}{2!}\) × 14 = \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!}\) × 14 = 25401600

Question 3.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8, without repetition.
Solution:
Here n = 5, r = 4, digits are {0, 2, 4, 7, 8}.
Hence, the sum of all 4 digited numbers that can be formed using the digits {0, 2, 4, 7, 8} without repetition is
^{(5 – 1)}P_{(4 – 1)} × (0 + 2 + 4 + 7 + 8) × (1111) – ^{(5 – 2)}P_{(4 – 2)} × (0 + 2 + 4 + 7 + 8) × (111)
= ⁴P₃ (21) × 1111 – ³P₂ (21) (111)
= 24 × 21 × 1111 – 6 (21) (111)
= 21 (26664) – 21 (666)
= 21 (26664-666)
= 21 (25998)
= 545958

Question 4.
Find the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4, 6, 8 without repetition.
Solution:
While forming any digit number With the given digits, zero cannot be filled in the first place.
We can fill the first place with the remaining 4 digits. The remaining places can be filled with the remaining 4 digits. All the numbers of 5 digits are greater than 4000. In the 4-digit numbers, the number starting with 4 or 6 or 8 are greater than 4000.

The number of 4-digit numbers that begin with 4 or 6 or 8 = 3 × ⁴P₃ = 3 × 24 = 72
The number of 5-digit numbers = 4 × 4! = 4 × 24 = 96
∴ The number of numbers greater than 4000 is 72 + 96 = 168.

III.

Question 1.
Find the number of ways of arranging 10 students A₁, A₂ …. A₁₀ in a row such that
(i) A₁, A₂, A₃ sit together.
(ii) A₁, A₂, A₃ sit in a specified order.
(iii) A₁, A₂, A₃ sit together in a specified order.
Solution:
A₁, A₂ …. A₁₀ are the ten students.
i) Consider A₁, A₂, A₃ as one unit and A₄, A₅, A₆, A₇, A₈, A₉, A₁₀ as seven units. These 8 units can be arranged in 8! ways.

The students A₁, A₂, A₃ in one unit can be arranged among themselves in 3! ways. The number of ways arranging 10 students in which A₁, A₂, A₃ sit together = (8!) (3!)

ii) To arrange A₁, A₂, A₃ to sit in a specified order,
A₁, A₂, A₃ can arrange in 10 positions in specific order in \(\frac{{ }^{10} \mathrm{P}_3}{3!}\) ways.

The remaining 7 people can be arranged in the remaining places in 7! ways.
∴ The number of ways of A₁, A₂, A₃ sit in a specific order = \(\frac{10!}{7!3!}\) × 7! = \(\frac{10!}{3!}\)

iii) To arrange A₁, A₂, A₃ sit together in a specified order.

Consider A₁, A₂, A₃ in that order as one unit. Now there are 8 objects, they can be arranged in 8! ways.
∴ The number of ways of A₁, A₂, A₃ sit together in specified order = 8! ways.

Question 2.
Find the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
Solution:
Here n = 5, r = 4, digits = {1, 2, 4, 5, 6}.
The sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition is
^(5-1)P_(4-1) × (1 + 2 + 4 + 5 + 6) × (1111)
= ⁴P₃ (18) (1111)
= 24 × 18 × 1111
= 4,79,952

Question 3.
9 different letters of an alphabet are given. Find the number of 4 letter words that can be formed using these 9 letters which have
(i) no letter is repeated
(ii) atleast one letter is repeated.
Solution:
The number of 4 letter words that can be formed from 9 different letters of an alphabet when repetition is allowed = 94 = 6561.
i) The number of 4 letter words the can be formed 9 different letters of an alphabet when no letter is repeated = ⁹P₄ = 9 × 8 × 7 × 6 = 3024.
ii) The number of 4 letter words that can be formed from 9 different letter of an alphabet with atleast one letter is repeated.
= 94 – ⁹P₄ = 6561 – 3024 = 3537.

Question 4.
Find the number of 4-digit numbers that can be formed using the digits 0, 1,2, 3, 4, 5 which are divisible by 6 when repetition of the digits is allowed.
Solution:
im-6
5 ways 6ways 7 ways 8 ways
Given digits 0, 1,2, 3, 4, 5 these are consecutive digits.
Total number of 4-digit numbers = 5 × 6³.
Among these total numbers some of the numbers are divisible by ‘6’.
By fixing first places with some fixed digits from 1, 2, 3, 4, 5 and 2nd, 3rd places with some fixed digits 0, 1, 2, 3, 4, 5 we get the following consecutive numbers.
a₁, a₂, a₃, 0
a₁, a₂, a₃, 1
a₁, a₂, a₃, 2
a₁, a₂, a₃, 3
a₁, a₂, a₃, 4
a₁, a₂, a₃, 5
We know that out of any 6 consecutive integers exactly 1 is divisble by 6.
∴ \(\frac{1}{6}\)th of these 5 × 6³ numbers are divisible by ‘6’.
Hence, required number = \(\frac{1}{6}\) × 5 × 6 = 5 × 36 = 180.

Question 5.
Find the number of ways of arranging the letters of the word ASSOCIATIONS. In how many of them
(i) all the three S’s come together
(ii) The two A’s do not come together
Solution:
The given word ASSOCIATIONS has 12 letters in which there are 2 As are alike, 3 Ss are alike, 2 O’s are alike, 2 I’s are alike and rest are different.
They can be arranged = \(\frac{12!}{2!3!2!2!}\)
i) Treat the 3 S’s as one unit. Then we have 9 + 1 = 10 entities in which there are 2 A’s are alike, 2 O’s are alike, 2 I’s are alike and rest are different.
They can be arranged in \(\frac{(10)!}{2!2!2!}\) ways.
The 3 S’s among themselves can be arranged in \(\frac{3!}{3!}\) = 1 way.
∴ The number of required arrangements = \(\frac{(10)!}{2!2!2!}\)

ii) Since 2 A’s do not come together, first arrange the remaining 10 letters in which there are 3. S’s are alike, 2 O’s are alike 2 I’s are alike and rest are different in \(\frac{(10)!}{3!2!2!}\) Ways.
Then we can find 11 gaps between them. The 2 A’s can be arranged in these 11 gaps in \(\frac{{ }^{11} \mathrm{P}_2}{2!}\) ways.
∴ The number of required arrangements = \(\frac{(10)!}{3!2!2!} \times \frac{{ }^{11} \mathrm{P}_2}{2!}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations Exercise 6b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6b

Question 1.
Evaluate (i) 8! (ii) 4! – 3!
Solution:
i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320
ii) 4! = 1 × 2 × 3 × 4 = 24 ⇒ 3! = 1 × 2 × 3 = 6
∴ 4! – 3! = 24 – 6 = 8

Question 2.
Is 3! + 4! = 7! ?
Solution:
3! = 1 × 2 × 3 = 6 and 4! = 1 × 2 × 3 × 4 = 24 ⇒ 3! + 4! = 6 + 24 = 30
But 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040; Hence, 3! + 4! ≠ 71.

Question 3.
Compute \(\frac{8!}{6!2!}\)
Solution:
\(\frac{8!}{6!2!}\)

= \(\frac{8 \times 7}{2}\)
= 28

Question 4.
If \(\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}\) find x.
Solution:
⇒ \(\frac{1}{6!} \times \frac{7}{7}+\frac{1}{7!}=\frac{x}{8!}\)

⇒ \(\frac{7}{7!}+\frac{1}{7!}=\frac{x}{8!}\)

⇒ \(\frac{8}{7!}\) × 8! = x

⇒ \(\frac{8}{7!}\) × 8 × 7! = x

⇒ x = 64

Question 5.
Evaluate \(\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!}\), when
i) n = 6, r = 2
Solution:
n = 6, r = 2
\(\frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}=\frac{6!}{4!}\)

= 30

ii) n = 9, r = 5
Solution:
n = 9, r = 5
\(\frac{n!}{(n-r)!}=\frac{9!}{(9-5)!}=\frac{9!}{4!}\)

= 9 × 8 × 7 × 6 × 5
= 15120

Referring to the AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations Exercise 6a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6a

Question 1.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
i) repetition of the digits is allowed ?
ii) repetition of the digits is not allowed?
Solution:
i) When repetition of the digits is allowed :
There will be as many ways as there are way’s of filling 3 vacant places

in succession by the given five digits.

In this case, repetition of digits is allowed.
Therefore, the unit’s place can be filled in by any of the given five digits.
Hence, by the multiplication principle, the number of ways in which three¬digit number can be formed from the given digits is 5 × 5 × 5 = 125.

ii) When repetition of the digits is not allowed :
In this case, repetition of digits is not allowed. Here, if unit place is filled the units place of the three digit number is 5.
Then, the ten’s place can be filled with any of the remaining four digit and the hundreds place can filled with any of the remaining three digits.
Hence, by the multiplication principle, the number of ways in which three¬digit number can be formed without repeating the given digits is 5 × 4 × 3 = 60.

Question 2.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated ?
Solution:
There will be as many ways as there are ways of filling 3 vacant places

in succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the unit’s place can be filled in 3 ways. The tens place, be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated.
Therefore, by multiplication principle, the required number of three digit even numbers are 3 × 6 × 6 = 108.

Question 3.
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated ?
Solution:
There are as many codes as there are way’s of filling 4 vacant places

in succession by the first 10 letters of the English alphabet, keeping in mind the repetition of letters is nqt allowed. The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet following which, the second place can be filled in be filled in by any of the remaining letters in 9 different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways.

Therefore, by multiplication principle.
The required numbers of ways in which 4 vacant places, can be filled is 10 × 9 × 8 × 7 = 5040. Hence, 5040 four letter codes can be formed using first 10 letters of the English alphabet, if no letter is repeated.

Question 4.
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once ?
Solution:
It is given that the 5-digit telephone numbers always start with 67.
Therefore, there will be as many phone numbers there are ways of filling 3 vacant places

by the digist 0 to 9, keeping in mind the digits 0 to 9, except digits 6 and 7.
Therefore, the unit’s place can be filled in 8 different ways following which, the ten’s place can be filled in by remaining 7 digits in 7 different ways, and the hundred’s place can filled in any of the remaining 6 digits in 6 different ways.
Therefore, by multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is 8 × 7 × 6 = 336.

Question 5.
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there ?
Solution:
When a coin is tossed, a different of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.
Hence, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2 = 8.

Question 6.
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Solution:
Each signals requires the use of 2 flags.
There will be as many flags as there are ways of filling in 2 vacant places

in succession by the given 5 flags of different colours.
The upper vacant place can be filled in 5 different ways by any one of 5 flags following which, the lower vacant place can be filled in 4 different ways by any one the remaining of different flags.
Thus, by multiplication principle, the number of different signals that can be generated is 5 x 4 = 20.

Referring to the AP Inter 1st Year Maths Study Material Chapter 5 Linear Inequalities Exercise 5d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Linear Inequalities Solutions Exercise 5d

Solve the following system of inequalities graphically.

Question 1.
x ≥ 3, y ≥ 2.
Solution:
Given that x ≥ 3 … (1)
y ≥ 2 …(2)
The graph of the lines, and x = 3 and y = 2, are drawn in the figure.
Inequality x ≥ 3 represents the region on the right hand side of the line, (including the line x = 2) and inequality y ≥ 2 represents the region above the line, (including the line y = 2).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2.
Solution:
Given that 3x + 2y ≤ 12 … (1)
x ≥ 1 ..(2)
y ≥ 2 … (3)
The graphs of the lines, 3x + 2y = 12, x = 1 and y = 2, are drawn in the figure. Inequality represents the region below the line 3x + 2y = 12 (including the line 3x + 2y = 12).

Inequality x ≥ 1 represents the region on the right hand side of the line x = 1 (including the line x= 1).
Inequality y ≥ 2 represents the region above the line, y = 2 (including the line y = 2).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 3.
2x + y ≥ 6, 3x + 4y ≤ 12.
Solution:
Given that 2x + y ≥ 6 … (1)
3x + 4y ≤ 12 …(2)
The graphs of the lines, 2x + y = 6 and 3x + 4y = 12 are drawn in the figure.

Inequality 2x + y ≥ 6 represents the region above the line, 2x + y = 6 (including the line 2x + y = 6).
Inequality 3x + 4y ≤ 12 represents the region below the line, 3x + 4y = 12 (including the line 3x + 4y = 12).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 4.
x + y ≥ 4, 2x – y < 0
Solution:
Given that x + y ≥ 4 … (1)
2x – y < 0 ….. (2)
The graphs of the lines, x + y = 4 and 2x – y = 0 are drawn in the figure. Inequality x + y > 4 represents the region above the line, x + y = 4 (including the line x + y = 4).
Inequality 2x – y < 0 represents the region above the line 2x – y = 0 (excluding the points on the line 2x-y = 0).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on the line 2x – y = 0 as shown in graph.

Question 5.
2x – y > 1, x – 2y < – 1.
Solution:
Given that 2x – y > 1 … (1)
x- 2y < -1 … (2)
The graphs of the lines, 2x – y = 1 and x – 2y = -1 are drawn in the figure.
Inequality 2x – y > 1 represents the region below the line, 2x – y = 1 (excluding the line 2x-y = 1).
Inequality x – 2y < – 1 represents the region above the line, x – 2y = -1 (excluding the line x – 2y = -1).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on the respective lines as shown in graph.

Question 6.
x + y ≤ 6, x + y ≥ 4.
Solution:
Given that x + y ≤ 6 … (1)
x + y ≥ 4 … (2)
The graphs of the lines, x + y = 6 and x + y = 4 and drawn in the figure.
Inequality x + y ≤ 6 represents the region below the line, x + y = 6 (including the line x + y = 6).
Inequality x + y ≥ 4 represents the region on the right hand side of the line, x + y = 4 (including the line x + y = 4).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10.
Solution:
Given that 2x + y ≥ 8 … (1)
x + 2y ≥ 10 … (2)
The graphs of the lines, 2x + y = 8 and x + 2y = 10 are drawn in the figure.
Inequality 2x + y ≥ 8 represents the region above the line,
2x + y = 8 (including the line).
Inequality x + 2y ≥ 10 represents the region on the right hand side of the line, x + 2y = 10 (including the line).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 8.
x + y ≤ 9, y > x, x ≥ 0.
Solution:
Given that x + y ≤ 9 … (1)
y > x …(2)
x ≥ 0 …(3)
The graph of the lines, x + y = 9 and y = x, are drawn in the figure.
Inequality x + y ≤ 9 represents the region below the line, x + y = 9 (including the line x + y = 9).
It is observed that (0, 1) satisfies the inequality, y > x. Since, 1 > 0.
Therefore, inequality y > x represents the half plane, corresponding to the line, y = x, containing the point (0,1), (excluding the line y = x).
Inequality x ≥ 0 represents the region on the right hand side of the line, x = 0 or Y-axis (including the Y-axis).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the lines, x + y = 9 and x = 0, but excluding the points on line y = x as shown in graph.

Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2.
Solution:
5x + 4y < 20 …(1)
x ≥ 1 …(2)
y ≥ 2 …(3)
The graph of the lines, 5x + 4y = 20, x = 1 and y = 2, are
Inequality 5x + 4y < 20 represents the region below the line, 5x + 4y = 20 (including the line 5x + 4y = 20).
Inequality x ≥ 1 represents the region on the right side of line, x = 1 (including the line x = 1).
Inequality y ≥ 2 represents the region above the line, y = 2 (including the line y = 2).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 10.
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:
Given that 3x + 4y ≤ 60 … (1)
x + 3y ≤ 30 …..(2)
x ≥ 0, y ≥ 0 … (3)
The graph of the lines, 3x + 4y = 60 and x + 3y = 30, are drawn in the figure.
Inequality 3x + 4y ≤ 60 represents the region below the line, 3x + 4y = 60 (including the line 3x + 4y = 60).
Inequality x +3y ≤ 30 represents the region below the line, x + 3y = 30 (including the line x + 3y = 30).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities as shown in graph.

Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6.
Solution:
Given that 2x + y ≥ 4 …(1)
x + y ≤ 3 …(2)
2x – 3y ≤ 6 …(3)
The graph of the lines, 2x + y = 4, x + y = 3 and 2x – 3y = 6, are drawn in the figure.
Inequality 2x + y ≥ 4 represents the region the line, 2x + y = 4 (including the line).
Inequality x + y ≤ 3 represents the region the line, x + y = 3 (including the line).
Inequality 2x – 3y ≤ 6 represents the region the line, 2x – 3y = 6 (including the line).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.
Solution:
Given that x – 2y ≤ 3 … (1)
3x + 4y ≥ 12 …(2)
x ≥ 0 …(3)
y ≥ 1 …(3)
The graphs of the lines, x – 2y = 3, 3x + 4y = 12, x = 0 and y = 1 are drawn in the figure.
Inequality x – 2y ≤ 3 represents the region above the line, x – 2y = 3 (including the line).
Inequality 3x + 4y ≥ 12 represents region above the line, 3x + 4y = 12 (including the line).
Inequality y ≥ 1 represents the region above the line, y = 1 (including the line).
Inequality x ≥ 0 represents the region on the right hand side of Y-axis (including Y-axis).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines and Y-axis as shown in graph.

Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3; x, y ≥ 0.
Solution:
Given that 4x + 3y ≤ 60 ………..(1)
y ≥ 2x, …………(2)
x ≥ 3; …………(3)
The graph of the lines, 4x + 3y = 60, y = 2x and x = 3, are drawn in the figure.
Inequality 4x + 3y ≤ 60 represents the region below the line, 4x + 3y = 60 (including the line 4x + 3y = 60).
Inequality y ≥ 2x represents the region above the line, y = 2x (including the line y = 2x).
Inequality x ≥ 3 represents the region on the right hand side of the line, x = 3 (including the line x = 3).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as shown in graph.

Question 14.
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0.
Solution:
3x + 2y ≤ 150 … (1)
x + 4y ≤ 80 … (2)
x ≤ 15 … (3)
The graph of the lines 3x + 2y = 150, x + 4y = 80 and x = 15, are drawn in the figure.
Inequality 3x + 2y ≤ 150 represents the region below the line, 3x + 2y = 150 (including the line 3x + 2y = 150).
Inequality x + 4y ≤ 80 represents the region below the line, x + 4y = 80 (including the line x + 4y = 80).
Inequality x ≤ 15 represents the region on the left hand side of the line, x = 15 (including the line x = 15).

Since x ≥ 1 and y ≤ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities as shown in graph.

Question 15.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0.
Solution:
Given that x + 2y ≤ 10 … (1)
x + y ≥ 1 …(2)
x – y ≤ 0 …(3)
The graph of the lines, x + 2y = 10, x + y = 1 and x – y = 0 are drawn in the figure.
Inequality x + 2y ≤ 10 represents the region below the line, x + 2y = 10 (including the line x + 2y = 10).
Inequality x + y ≥ 1 represents the region above the line, x + y = 1 (including the line x + y = 1).
Inequality x – y ≤ 0 represents the region above the line, x – y = 0 (including the line x – y = 0).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities as shown in graph.

Referring to the AP Inter 1st Year Maths Study Material Chapter 5 Linear Inequalities Exercise 5c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Linear Inequalities Solutions Exercise 5c

Solve the following inequalities graphically in two-dimensional plane,

Question 1.
x + y < 5.
Solution:
The graphical representation of x + y = 5 is given as dotted line in the figure. This line divides the xy-plane in two planes I and II.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.
We select the point as (0, 0).

It is observed that,
0 + 0 < 5 or 0 < 5, which is true.
Therefore, half-plane II is not the solution region of the given inequality. Also, it is evident that any point on the line does not satisfy the given strict inequality.
Thus, the solution region of the given inequality is the shaded half plane I excluding the points on the line.
This can be represented as shown in graph.

Question 2.
2x + y ≥ 6.
Solution:
The graphical representation of 2x + y = 6 is given in the figure. This line divides the xy-plane in two half planes, I and II.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 2(0) + 0 ≥ 6 or 0 ≥ 6, which is false.
Therefore, half plane I is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality.
Thus, the solution region of the given inequality is the shaded half plane II including the points on the line.
This can be represented as shown in graph.

Question 3.
3x + 4y ≤ 12.
Solution:
The graphical representation of 3x + 4y = 12 is given in the figure. This line divides the xy-plane in two half planes, I and II.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 3(0) + 4(0) < 12 or 0 < 12, which is true.
Thus, the solution region of the given in equality is the shaded half plane I, including the points on the line.
This can be represented as shown in graph.

Question 4.
y + 8 ≥ 2x.
Solution:
The graphical representation of y + 8 = 2x is given in the figure. This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 0 + 8 ≥ 2(0) or 8 ≥ 0, which is true.
Therefore, lower half plane is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality.
Thus, the solution region of the given inequality is the upper half plane containing the points including the line.
The solution region is represented by the shaded region as shown in graph.

Question 5.
x – y ≤ 2.
Solution:
The graphical representation of x – y = 2 is given in the figure. This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 0 – 0 ≤ 2 or 0 ≤ 2, which is true.
Therefore, the lower half plane is not the solution region of the given inequality. Also, any point on the line satisfies the given inequality.
Thus, the solution region of the given inequality is the upper half plane containing the points including the line.
The solution region is represented by the shaded region as shown in graph.

Question 6.
2x – 3y > 6.
Solution:
The graphical representation of 2x – 3y = 6 is given as dotted line in the figure. This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 2(0) – 3(0) ≥ 6 or 0 ≥ 6, which is false.
Therefore, the upper half plane is not the solution region of the given inequality. Also, any point on the line does not satisfy the given inequality.
Thus, the solution region of the given inequality is the lower half plane that does not contain the points including the line.
The solution region is represented by the shaded region as shown in graph.

Question 7.
– 3x + 2y ≥ – 6.
Solution:
The graphical representation of -3x + 2y = -6 is given in the figure. This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that -3(0) + 2(0) ≥ -6 or 0 ≥ – 6, which is true. .
Therefore, the lower half plane is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality.
Thus, the solution region of the given inequality is the upper half plane containing the points including the line.
The solution region is represented by the shaded region as shown in graph.

Question 8.
3y – 5x < 30.
Solution:
The graphical representation of 3y – 5x = 30 is given as dotted line in the figure. This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 3(0) – 5(0) < 30 or 0 < 30, which is true.
Therefore, the upper half plane is not the solution region of the given inequality. Also, it is evident that any point on the line does not satisfy the given inequality.
Thus, the solution region of the given inequality is the lower half plane containing the points excluding the line.
The solution region is represented by the shaded region as shown in graph.

Question 9.
y < -2.
Solution:
The graphical representation of y = – 2 is given as dotted line in the figure. This line divides the xiy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 0 < -2, which is false.
Also, it is evident that any point on the line does not satisfy the given inequality.
Hence, every point below the line, y = – 2 (excluding all the points on the line), determine the solution of the given inequality.
The solution region is represented by the shaded region as shown in graph,

Question 10.
x > – 3.
Solution:
The graphical representation of x = – 3 is given as dotted line in the figure. This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.

We select the point as (0, 0).
It is observed that, 0 > -3, which is true.
Also, it is evident that any point on the line does not satisfy the given inequality.
Hence, every point on the right side of the line, x= -3 (excluding all the points on the line), determines the solution of the given inequality.

Practice AP Inter 1st Year Maths Study Material Chapter 5 Linear Inequalities MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Linear Inequalities MCQ

Question 1.
The length of a rectangle is four times its breadth. If the minimum perimeter of the rectangle is 160 cm, then
1) breadth >16 cm
2) length < 20 cm
3) breadth ≥ 16 cm
4) length < 20 cm
Answer:
3) breadth >16 cm

Explantaion:
Let x be the length and y be the breadth of the rectangle.
Given x = 4y
Perimeter of the rectangle 2(x + y) ≥ 160
⇒ x + y ≥ 80
⇒ 4y + y ≥ 80
⇒ 5y ≥ 80
⇒ y ≥ 16

Question 2.
Consider a linear inequality -3x + 15 < -12, then
1) x ∈ (9, ∞)
2) x ∈ [9, ∞)
3) x ∈ (-∞, 9]
4) x ∈ [-9, 10)
Answer:
1) x ∈ (9, ∞)

Explantaion:
Given -3x + 15 < -12
⇒ 15 + 12 < 3x
⇒ 3x > 27
⇒ x > 9
⇒ x ∈ (9, ∞)

Question 3.
Suppose we have three real numbers which include x, y and b given by x > y, b < 0. Determine the relation between the three variables.
1) x/b < y
2) x ≤ y
3) x/b < y/b
4) x ≥ y/b
Answer:
3) x/b < y/b

Explantaion:
\(\frac{x}{b}<\frac{y}{b}\)

Question 4.
If | x – 2 | > 6, then
1) x ∈ (- 4, 8)
2) x ∈ [-4, 8]
3) x ∈ (-∞, – 4) ∪ (8, ∞)
4) x ∈ (-∞, -4) ∪ [8, ∞)
Answer:
3) x ∈ (-∞, – 4) ∪ (8, ∞)

Explantaion:
|x – 2| > 6
x – 2 < -6 (or) x – 2 > 6
⇒ x < -4 or x > 8
⇒ x ∈ (-∞, – 4) ∪ (8, ∞)

Question 5.
If \(\frac{|x-8|}{(x-8)}\) ≥ 0, then
1) x ∈ [8, ∞)
2) x ∈ (7, 8)
3) x ∈ (-∞,8)
4) x ∈ (8, ∞)
Answer:
4) x ∈ (8, ∞)

Explantaion:
For the given expression, it can be defined as
x – 8 ≠ 0 ⇒ x ≠ 8
(x – 8) > 0 ⇒ x > 8
⇒ x ∈ (8, ∞)

Question 6.
If |x + 5| ≥ 10, then
1) x ∈ (-15,15]
2) x ∈ (-15, 5]
3) x ∈ (-∞, -15] ∪ (5, ∞)
4) x ∈ (-∞, -15] ∪ [15, ∞)
Answer:
3) x ∈ (-∞, -15] ∪ (5, ∞)

Explantaion:
|x + 5| > 10
⇒ ±(x + 5) ≥ 10
⇒ (x + 5) ≥ 10 (or) -(x + 5) ≥ 10
⇒ x ≥ 5 (or) x ≤ -15
⇒ x ∈ (-∞, -15] ∪ (5, ∞)

Question 7.
If 4x + 4 < 6x + 5, then x belongs to the interval
1) (2, ∞)
2) (-\(\frac{1}{2}\), ∞)
3) (-∞, \(\frac{1}{2}\))
4) (-4, ∞)
Answer:
2) (-\(\frac{1}{2}\), ∞)

Explantaion:
4x + 4 < 6x + 5
⇒ 6x – 4x > 4 – 5
⇒ 2x > -1
⇒ x > –\(\frac{1}{2}\)
⇒ x ∈ (-\(\frac{1}{2}\), ∞)

Question 8.
The solution of the linear inequality 2x – 17 > 3 – 8x is
1) x ≥ 2
2) x < 2
3) x ≤ 2
4) x > 2
Answer:
4) x > 2

Explantaion:
2x – 17 > 3 – 8x
⇒ 2x + 8x > 3 + 17
⇒ 10x > 20
⇒ x > 2

Question 9.
Isha scored 90 and 90 marks in the first two-class tests. The minimum marks she should get in the third test to have an average of atleast 90 marks, is
1) x ≥ 90
2) x > 90
3) x ≤ 90
4) x < 90
Answer:
1) x ≥ 90

Explantaion:
Let x be the minimum marks
Average marks of Isha to get an average of atleast 90
⇒ \(\frac{1}{2}\) ≥ 90
⇒ 180 + x ≥ 270
⇒ x ≥ 90

Question 10.
The set of values of x in 5x + 2 > 12 is (x is a real number)
1) (-2, 2)
2) (-1, ∞)
3) (2, ∞)
4) (-2, 1)
Answer:
3) (2, ∞)

Explantaion:
5x + 2 > 12
⇒ 5x > 12 – 2
⇒ 5x > 10
⇒ x > 2
⇒ x ∈ (2, ∞)

Question 11.
When x is an integer, the set of values of x satisfying the linear inequality
5x – 2 ≤ 3x + 2 are
1) {……….. -2, -1, 0, 1}
2) {…,-3,-2,-1, 0, 1}
3) {…, -2, -1, 0, 1, 2}
4) {…, -2, -1, 0, 1, 2, 3}
Answer:
3) {…, -2, -1, 0, 1, 2}

Explantaion:
5x – 2 ≤ 3x + 2
⇒ 5x – 3x ≤ 2 + 2
⇒ 2x ≤ 4
⇒ x ≤ 2
∴ x = {…,-2, -1, 0, 1, 2}

Question 12.
The set of values of x satisfying 50x ≤ 150, where x is the natural numberis
1) {1, 2, 3, 4}
2) {1, 2}
3) {0, 1, 2}
4) {1, 2, 3}
Answer:
4) {1, 2, 3}

Explantaion:
50x ≤ 150
⇒ x ≤ 3
⇒ x ∈ {1, 2, 3}

Question 13.
If 9x > -36, then the least integral value of x is
1) -3
2) -4
3) -5
4) -6
Answer:
1) -3

Explantaion:
9x > -36
⇒ x > – 4
∴ x is valid for – 3