Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(e) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(e)

I.

Question 1.

If \({ }^{\mathrm{n}} \mathrm{C}_4\) = 210, find n.

Solution:

Given \({ }^{\mathrm{n}} \mathrm{C}_4\) = 210

⇒ \(\frac{n(n-1)(n-2)(n-3)}{4 !}\) = 210

⇒ n (n – 1) (n – 2) (n – 3) = 41 × 210

= 24 × 210

= 7 × 8 × 9 × 10

On comparing largest integers, we get n = 10.

Question 2.

If \({ }^{12} \mathrm{C}_{\mathrm{r}}\) = 495, find the possible values of ‘r’.

Solution:

Given \({ }^{12} \mathrm{C}_{\mathrm{r}}\) = 495

= 11 × 9 × 5

= \(\frac{12 \times 11 \times 9 \times 10}{4 \times 3 \times 2 \times 1}\)

⇒ \({ }^{12} \mathrm{C}_{\mathrm{r}}={ }^{12} \mathrm{C}_4 \text { or }{ }^{12} \mathrm{C}_8\)

⇒ r = 4 or 8.

Question 3.

If 10 . \({ }^n \mathrm{C}_2\) = 3 . \({ }^{n+1} C_3\), find n.

Solution:

Given 10 . \({ }^n C_2\) = 3 . \({ }^{n+1} C_3\)

\(10 \times \frac{n(n-1)}{2}=3 \cdot \frac{(n+1) n(n-1)}{3 \times 2 \times 1}\)

⇒ 10 = n + 1

⇒ n = 9.

Question 4.

If \({ }^n P_r\) = 5040 and \({ }^n C_r\) = 210, find n and r.

Solution:

Given \({ }^n P_r\) = 5040 and \({ }^n C_r\) = 210, \({ }^n P_r=r !^n C_r\)

5040 = r! × 210

⇒ r! = 24

⇒ r! = 4!

∴ r = 4

∴ \({ }^n \mathrm{P}_4\) = 5040

⇒ n (n – 1) (n – 2) (n – 3) = 10 × 9 × 8 × 7

On comparing largest integers, we get n = 10

∴ n = 10 and r = 4.

Question 5.

If \({ }^n C_4={ }^n C_6\), find n.

Solution:

Given \({ }^n C_4={ }^n C_6\)

If \({ }^n C_r={ }^n C_s\), then either r = s or r + s = n.

Clearly, we have n = 4 + 6

⇒ n = 10.

Question 6.

If \({ }^{15} \mathrm{C}_{2 \mathrm{r}-1}={ }^{15} \mathrm{C}_{2 \mathrm{r}+4}\), find r.

Solution:

Given \({ }^{15} \mathrm{C}_{2 \mathrm{r}-1}={ }^{15} \mathrm{C}_{2 \mathrm{r}+4}\)

If \({ }^n C_r={ }^n C_s\) then either r = s or r + s = n.

∴ 2r – 1 = 2r + 4

Which is impossible.

or

2r – 1 + 2r + 4 = 15

⇒ 4r + 3 = 15

⇒ r = 3

∴ r = 3.

Question 7.

If \({ }^{17} C_{2 t+1}={ }^{17} C_{3 t-5}\), find t.

Solution:

Given \({ }^{17} C_{2 t+1}={ }^{17} C_{3 t-5}\)

If \({ }^n C_r={ }^n C_s\), then either r = s or n = r + s.

i.e., either

2t + 1 = 3t – 5

⇒ t = 6 = 17

2t + 1 + 3t – 5 = 17

⇒ 5t = 21

⇒ t = \(\frac{21}{5}\)

Since t’ is an integer, we have t = 6.

Question 8.

If \({ }^{12} C_{r+1}={ }^{12} C_{3 r-5}\), find r.

Solution:

Given \({ }^{12} C_{r+1}={ }^{12} C_{3 r-5}\).

If \({ }^n C_r={ }^n C_s\), then either r = s or n = r + s.

i.e., r + 1 = 3r – 5

or 12 = r + 1 + 3r – 5

⇒ r = 3 or r = 4.

Question 9.

If \({ }^9 C_3+{ }^9 C_5={ }^{10} C_r\), then find r.

Solution:

Given \({ }^9 C_3+{ }^9 C_5={ }^{10} C_r\)

⇒ \({ }^9 \mathrm{C}_3+{ }^9 \mathrm{C}_4={ }^{10} \mathrm{C}_r\) (∵ \({ }^n C_r={ }^n C_s\))

⇒ \({ }^{10} \mathrm{C}_4={ }^{10} \mathrm{C}_{\mathrm{r}}\) (or) \({ }^{10} \mathrm{C}_6={ }^{10} \mathrm{C}_{\mathrm{r}}\)

⇒ r = 4 or r = 6.

Question 10.

Find the number of ways of forming a com-mittee of 5 members from 6 men and 3 ladies.

Solution:

Selecting 5 members to form a commitee from 6 men and 3 ladies (i.e., 9 members) can be done in \({ }^9 \mathrm{C}_5\) = 126 ways.

Question 11.

In question 10, how many committees contain atleast two ladies.

Solution:

In selecting 5 members from 6 men and 3 ladies to form a committee containing atleast two ladies, two cases arises.

Case – (1):

(When committee contains exactly two ladies) :

Number of ways of selecting 2 ladies from 3 ladies is \({ }^3 \mathrm{C}_2\).

Now the remaining 3 members are selected from 6 men and this can be done in C3 ways

∴ Number of ways to form a committee with 2 ladies = \({ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_3\) = 60.

Case – (2)

(When committee contains 3 ladies) :

Selecting 3 ladies from 3 ladies can be done in \({ }^3 \mathrm{C}_3\) ways.

Selecting remaining 2 members from 6 men can be done in \({ }^6 \mathrm{C}_2\) ways.

∴ Number of ways to form a committee with 3 ladies = \({ }^3 \mathrm{C}_3 \times{ }^6 \mathrm{C}_2\) = 15

∴ Total number of ways = 60 + 15 = 75.

Question 12.

If \({ }^n C_5={ }^n C_6\), then \({ }^{13} \mathrm{C}_{\mathrm{n}}\).

Solution:

Given, \({ }^n C_5={ }^n C_6\)

⇒ n = 5 + 6

(If \({ }^n C_r={ }^n C_s\) then either n = r + s or r = s)

⇒ n = 11

Now, \({ }^{13} C_n={ }^{13} C_{11}={ }^{13} C_2\) = 78.

II.

Question 1.

Prove that 3 ≤ r ≤ n, \({ }^{(n-3)} C_r+3 \cdot{ }^{(n-3)} C_{(r-1)}+3 \cdot{ }^{(n-3)} C_{(r-2)}+{ }^{(n-3)} C_{(r-3)}={ }^n C_r\).

Solution:

Given 3 ≤ r ≤ n

Question 2.

Find the value of \({ }^{10} C_5+2 \cdot{ }^{10} C_4+{ }^{10} C_3\).

Solution:

\({ }^{10} C_5+2 \cdot{ }^{10} C_4+{ }^{10} C_3\)

= \({ }^{10} C_5+{ }^{10} C_4+{ }^{10} C_4+{ }^{10} C_3\)

= \({ }^{11} C_5+{ }^{11} C_4\) (∵ \({ }^n C_{r-1}+{ }^n C_r={ }^{n+1} C_r\))

= \({ }^{12} \mathrm{C}_5\) = 792.

Question 3.

Simplify \({ }^{34} C_5+\sum_{r=0}^4(38-r) C_4\).

Solution:

Question 4.

In a class there are 30 students. If each student plays a chess gaine 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.

Given each students plays a chess game with each of the other student.

∴ The total number of chess games played is equal to number of ways of selecting 2 students to play a game from 30 students.

This can be done in \({ }^{30} \mathrm{C}_2\) ways.

∴ The number of chess games played = \({ }^{30} \mathrm{C}_2\) = 435.

Question 5.

Find the number of ways of selectIng 3 girls and 3 boys out of 7 girls and 6 boys.

Solution:

Number of ways of selecting 3 girls out of 7 girls = \({ }^7 \mathrm{C}_3\)

Number of ways of selecting 3 boys ouf of 6 boys = \({ }^6 \mathrm{C}_3\)

∴ The total number of ways = \({ }^7 \mathrm{c}_3 \cdot{ }^6 \mathrm{c}_3\)

= 35 . 20 = 700.

Question 6.

Find the number of ways of selecting a committee of 6 members out of 10 mem bers always Including a specified member.

Solution:

A committee of 6 members is to be formed out of 10 members in which a specified member is always included.

So, remaining 5 members are to be selected from rest of 9 members.

This can be done in \({ }^9 \mathrm{C}_5\) ways.

∴ Required number of ways \({ }^9 \mathrm{C}_5\) = 126.

Question 7.

Find the number of ways of selecting 5 books from 9 different mathematics books such that a particular book ¡s not included.

Solution:

Given out of 9 different mathematics books a particular book is not included.

∴Number of book left are ‘8′.

∴ Number of ways of selecting 5 books out of 8 different books are \({ }^8 C_5\) = 56.

Question 8.

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.

Number of ways of selecting 3 vowels out of 5 = \({ }^5 \mathrm{C}_3\) = 10

Number of ways of selecting 2 consonants out of 3 = \({ }^3 \mathrm{C}_2\) = 3

∴ Total number of ways = 10 x 3 = 30.

Question 9.

Find the number of diagonals of a polygon with 12 sides.

Solution:

Number of sides of a polygon = 12

Number of diagonals of a n – sided polygon = \({ }^n C_2\) – n

∴ Number of diagonals of 12 sided polygon = \({ }^{12} C_2\) – 12 = 54.

Question 10.

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:

Number of ways of selecting 2 persons out of n persons sitting in a row, who are sitting adjacent to each other = n – 1.

Question 11.

Find the number of ways of giving away 4 similar coins to 5 boys if each boy can be given any number (less than or equal to 4) of coins.

Solution:

In distribution of 4 similar coins to 5 boys, the following cases arises.

Case – (i) :

Giving all 4 coins to one boys. This is done in \({ }^5 \mathrm{C}_1\) ways.

Case – (ii) :

Giving 4 coins to two boys so that one of them gets 1 and the other 3 coins.

This is done in 2 x \({ }^5 \mathrm{C}_2\) ways.

Case – (iii) :

Giving 4 coins to two boys so that each get 2 coins. This can be done in \({ }^5 \mathrm{C}_2\) ways.

Case – (iv) :

Giving 4 coins to three boys so that, two of them gets 1 coin and the other gets 2. This is done in \({ }^5 \mathrm{C}_3 \times \frac{3 !}{2 !}\) ways.

Case – (v):

Giving 4 coins to four boys so that each gets 1.

This is done in \({ }^5 \mathrm{C}_4\) ways.

∴ Total number of ways = \({ }^5 \mathrm{C}_1+2 \times{ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_2+\frac{3 !}{2 !}{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4\) = 70.

III .

Question 1.

Prove that \(\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1 \cdot 3 \cdot 5 \ldots \ldots(4 n-1)}{\{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)\}^2}\).

Solution:

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:

Given number of elements in set A are 12.

i) Subsets of A having 4 elements :

Number of subsets of A having 4 elements is equal to number of ways of selecting 4 elements from 12 elements in set ‘A’.

This can be done in \({ }^{12} C_4\) ways.

∴ Number of subsets of A having 4 elements = \({ }^{12} C_4\) = 495.

ii) Subset of A contains atleast 3 elements:

Number of subsets of A, having ‘r’ elements is equal to number of ways of selecting ‘r’ elements from 12 elements in set A’, i.e., \({ }^{12} \mathrm{C}_{\mathrm{r}}\) ways.

∴ Number of ways of selecting at least 3 elements from 12 elements in set A is \({ }^{12} \mathrm{C}_3+{ }^{12} \mathrm{C}_4+\ldots \ldots+{ }^{12} \mathrm{C}_{12}\)

Number of subsets of A having atleast 3 elements = \({ }^{12} \mathrm{C}_3+{ }^{12} \mathrm{C}_4+\ldots \ldots+{ }^{12} \mathrm{C}_{12}\)

= \(\left({ }^{12} \mathrm{C}_0+{ }^{12} \mathrm{C}_1+\ldots \ldots+{ }^{12} \mathrm{C}_{12}\right)-{ }^{12} \mathrm{C}_0-{ }^{12} \mathrm{C}_1-{ }^{12} \mathrm{C}_2\)

= 2^{12} – \({ }^{12} \mathrm{C}_0+{ }^{12} \mathrm{C}_1+{ }^{12} \mathrm{C}_2\) = 4017.

iii) Number of subsets of ‘A’ having atmost 3 elements :

Number of subsets of A having ‘r’ elements is equal to number of ways of selecting r’ elements from 12 elements in set ‘A’ i.e., \({ }^{12} C_r\) ways.

∴ Number of ways of selecting atmost 3 elements from 12 elements in set A is \({ }^{12} \mathrm{C}_0+{ }^{12} \mathrm{C}_1+{ }^{12} \mathrm{C}_2+{ }^{12} \mathrm{C}_3\).

∴ Number of subsets of ‘A’ having atmost 3 elements = \({ }^{12} \mathrm{C}_0+{ }^{12} \mathrm{C}_1+{ }^{12} \mathrm{C}_2+{ }^{12} \mathrm{C}_3\) = 299.

Question 3.

Find the numbers of ways of selecting a cricket team of 11 players from 7 batsmen and 6 bowlers such that there will be atleast 5 bowlers in the team.

Solution:

Number of batsmen = 7

Number of bowlers = 6

In selecting 11 players in a team out of given 13 players so that the team contains atleast 5 bowlers, two cases arises.

Case (i) : (Selecting 5 bowlers) :

Number of ways of selecting 5 bowlers from 6 = \({ }^6 \mathrm{C}_5\)

The remaining 6 players are selected from 7 batsmen can be done in \({ }^7 \mathrm{C}_6\) ways.

Number of ways of selecting = \({ }^7 \mathrm{C}_6 \times{ }^6 \mathrm{C}_5\).

Case – (ii) (Selecting 6 bowlers) :

Number of ways of selecting 6 bowlers from 6 = \({ }^6 \mathrm{C}_6\)

The remaining 5 players to be selected from 7 batsmen can be done in \({ }^7 \mathrm{C}_5\) ways.

∴ Number of ways of selecting = \(\mathrm{C}_6 \times{ }^7 \mathrm{C}_5\)

Total number of ways of selecting = \({ }^7 \mathrm{C}_6 \times{ }^7 \mathrm{C}_5+{ }^7 \mathrm{C}_5 \times{ }^6 \mathrm{C}_6\) = 63.

Question 4.

In 5 vowels and 6 consonants are given, then how many 6 letter words can be formed with 3 vowels and 3 consonants.

Solution:

Given 5 vowels and 6 consonants.

6 letter word is formed with 3 vowels and 3 consonants.

Number of ways of selecting 3 vowels from 5 vowels is \({ }^5 \mathrm{C}_3\).

Number of ways of selecting 3 consonants from 6 consonants is \({ }^6 \mathrm{C}_3\).

∴ Total number of ways of selecting = \({ }^5 \mathrm{C}_3 \times{ }^6 \mathrm{C}_3\)

These letters can be arranged themselves in 6! ways.

∴ Number of 6 letter words formed = \({ }^5 \mathrm{C}_3 \times{ }^6 \mathrm{C}_3\) × 6!.

Question 5.

There are 8 railway stations along a rail-way line. In how many ways can a train be stopped at 3 of these stations such that no two of them are consecutive ?

Solution:

Let S_{1}, S_{2}, S_{3} ……. S_{8} be 8 railway stations along a railway line.

Train is to be stopped at 3 stations.

Number of ways of selecting 3 stations out of 8 stations is 8Cr

Number of ways of selecting 3 consecutive stations is 6.

(i.e., (S_{1}, S_{2}, S_{3}), (S_{2}, S_{3}, S_{4}), ………. (S_{6}, S_{7}, S_{8})}

Number of ways of selecting only 2 consecu¬tive stations = 2 × 5 + 5 × 4 = 30

As no two stops are consecutive, number of ways of selecting = \({ }^8 \mathrm{C}_3\) – 6 – 30 = 20.

Question 6.

Find the number of ways of forming a com¬mittee of 5 members out of 6 Indians and 5 Americans so that always the Indians will be in majority in the committee.

Solution:

A committee of 5 members is to be formed out of 6 Indians and 5 Americans.

As committee contains the majority of Indians, 3 cases arises.

i) Selecting 3 Indians and 2 Americans :

Number of ways of selecting 3 Indians out of 6 Indians = \({ }^6 \mathrm{C}_3\)

Number of ways of selecting 2 Americans out of 3 Indians = \({ }^5 \mathrm{C}_2\)

Number of ways of selecting 3 Indians and 2 Americans = \({ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2\).

ii) Selecting 4 Indians and 1 American :

Number of ways of selecting 4 Indians out of 6 Indians = \({ }^6 \mathrm{C}_4\)

Number of ways of selecting 1 American out of 5 Americans = \({ }^5 \mathrm{C}_1\)

Number of ways of selecting 4 Indians and 1 American = \({ }^6 \mathrm{C}_4 \times{ }^5 \mathrm{C}_1\).

iii) Selecting 5 Indians :

Number of ways of selecting all 5 members

Indians out of 6 Indians = \({ }^6 \mathrm{C}_5\).

∴ Total numbers of ways of forming a committee = \({ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2+{ }^6 \mathrm{C}_4 \times{ }^5 \mathrm{C}_1+{ }^6 \mathrm{C}_5\) = 281.

Question 7.

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:

A question paper contains 3 sections A, B, C containing 3, 4, 5 questions respectively.

Number of ways of selectng 6 questions out of these 12 questions = \({ }^{12} \mathrm{C}_6\)

Number of ways of selecting 6 questions from sections B and C (i.e., from 9 questions) = \({ }^{9} \mathrm{C}_6\)

Number of ways of selecting 6 questions from sections A and C (i.e., from 8 questions) = \({ }^{8} \mathrm{C}_6\)

Number of ways of selecting 6 questions from sections A and B (i.e., 7 questions) = \({ }^{7} \mathrm{C}_6\)

∴ Number of ways of selecting 6 questions choosing atleast one from each section = \({ }^{12} \mathrm{C}_6-{ }^7 \mathrm{C}_6-{ }^8 \mathrm{C}_6-{ }^9 \mathrm{C}_6\) = 805.

Question 8.

Find the number of ways in which 12 things be

(i) divided into 4 equal groups

(ii) distributed to 4 persons equally.

Solution:

i) Dividing 12 things in 4 equal groups :

Number of ways of dividing 12 things into 4 equal groups = \(\frac{12 !}{(3 !)^4 \cdot 4 !}\).

ii) Distributing 12 things to 4 persons equally

Number of ways of distributing 12 things to 4 persons equally = \(\frac{12 !}{(3 !)^4 \cdot 4 !}\).

Question 9.

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:

A class contains 4 boys and ‘g’ girls,

In selecting 5 students with atleast 3 boys for picnic two cases arises.

i) Selecting 3 boys and 2 girls :

Number of ways of selecting 3 boys and 2 girls = \({ }^4 C_3 \times{ }^g C_2=4\left({ }^g C_2\right)\)

As each group contains 2 girls, number of dolls required = 8 \(8\left({ }^8 \mathrm{C}_2\right)\).

ii) Selecting 4 boys and 1 girl :

Number of ways of selecting 4 boys and 1 girl = \({ }^4 \mathrm{C}_4 \times{ }^{\mathrm{g}} \mathrm{C}_1\) = g

∴ As each group contains only 1 girl, number of dolls required = g

∴ Total number of dolls = 8 (\(\left({ }^g \mathrm{C}_2\right)\)) + g

i.e., 85 = \(\frac{g(g-1)}{2}\) + g

⇒ 85 = 4g^{2} – 3g

⇒ 4g^{2} – 3g – 85 = 0

⇒ (4g + 17) (g – 5) = 0

⇒ g = 5 (∵ ‘g’ is non-negative integer).