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

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

I.

Question 1.

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:

Given digits are 1, 2, 4, 5, 7, 8.

As repetitions are allowed,

Each place of 4 – digit number can be filled by given ‘6’ digits in 6 ways.

∴ By fundamental principle of counting number of 4 – digit numbers are 6 × 6 × 6 × 6 = 6^{4} = 1296.

Question 2.

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:

Given word RHYME contains 5 different letters.

As repetitions are allowed, each blank of 5 letter words that can be formed using letters of word RHYME is 5 × 5 × 5 × 5 × 5 = 5^{5} = 3125.

Question 3.

Find the number of functions from a set A containing 5 elements into a set B containing 4 elements.

Solution:

Let A = {a_{1}, a_{2}, a_{3}, a_{4}, a_{5}} and B = {b_{1}, b_{2}, b_{3}, b_{4}}

To define the image of a, we have 4 choices in set B.

i. e., Each element of set A has 4 choices in set B.

∴ The number of functions from A to B is 4 × 4 × 4 × 4 × 4 = 4^{5} = 1024.

II.

Question 1.

Find the number of palindromes with 6 digits that can be formed using the digits

i) 0, 2, 4, 6, 8

ii) 1, 3, 5, 7, 9.

Solution:

i) Given digits are 0, 2, 4, 6, 8.

The first place and last place (lakh’s place and unit’s place) of a 6 – digit palindrome number is filled by same digit.

This can be done in 4 ways (using non¬zero digit).

Similarly Ten thousand’s place and ten’s place is filled by same digit.

As repetition is allowed this can be done in 5 ways.

Thousand’s place and Hundred’s place is filled by same digit in 5 ways.

∴ Total number of 6 digital palindromes formed using given digits are 4 × 5 × 5 = 100.

ii) Given digits are 1, 3, 5, 7, 9.

The first place and last place (i.e., lakh’s place and unit’s place) of a 6 – digit palindrome number is filled by same digit in 5 ways.

As repetitions allowed,

Similarly ten thousand’s place & ten’s place is filled by same digit in 5 ways.

Thousand’s place and Hundred’s place is filled by same digit in 5 ways.

Total number of 6 digit palindrome formed using given digits are 5 × 5 × 5 = 125.

Question 2.

Find the number of 4 – digit telephone numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 with atleast one digit repeated.

Solution:

Given digits are 1, 2, 3, 4, 5, 6.

The number of 4-digit telephone numbers that can be formed using the given 6 digits.

Case – (i) :

When repetitions is allowed = 6^{4}

Case – (ii) :

When repetitions is not allowed

Hence the number of 4 digit telephone numbers in which atleast one digit repeated is 6^{4} – \({ }^6 \mathrm{P}_4\) = 936.

Question 3.

Find the number of bijections from a set A containing 7 elements onto itself.

Solution:

Let A = {a7, a2 a7) i.e., set containing 7 elements.

The bijection is both one-one and onto.

So, to define the image of a_{1} we have 7 choices.

Then we can define the image of a_{2} is 6 ways.

Similarly we can define the image of a_{3} in 5 ways.

Proceeding like this, the image of a_{7} is defined only in one way.

∴ The number of bijections from A onto A is 7 × 6 × 5 × …………… × 1 = 7! = 5040.

Question 4.

Find the number of ways of arranging ‘r’ things in a line using the given ‘n’ different things in which atleast one thing is repeated.

Solution:

The number of ways of arranging r’ things in a line using the given n’ different things, when repetitions is allowed is n^{r}.

The number of ways of arranging ‘r’ things in a line using the given n’ different things. When repetitions is not allowed is \({ }^n P_r\).

∴ The number of ways of arranging ‘r’ things in a line using n’ different things so that atleast one thing is repeated is n^{r} – \({ }^n P_r\).

Question 5.

Find the number of 5 letter words that can be formed using the letters of the word NATURE that begin with N when repetition is allowed.

Solution:

Given word is ‘NATURE’.

As the 5 letter word begins with ‘N’, the first place is filled in only 1 way.

As repetitions is allowed, each place of remaining 4 places can be filled in 6 ways.

∴ Total number of 5 letter words formed are 6^{4} = 1296.

Question 6.

Find the number of 5-digit numbers divis-ible by 5 that can be formed using the digits 0, 1, 2, 3, 4, 5, when repetition is allowed.

Solution:

Given digits are 0, 1, 2, 3, 4, 5.

The ten thousand’s place of a 5 – digit numbers formed using given digits can be filled in 5 ways.

As the 5 – digit number is divisible by ‘5’, the unit’s place can be filled in 2 ways.

(i.e., either 0 or 5).

∴ The remaining 3 places can be filled in 6 ways each.

∴ Number of 5 digit numbers divisible by 5 formed using given digits = 5 × 2 × 6^{3} = 2160.

Question 7.

Find the number of numbers less than 2000 that can be formed using the digits, 1, 2, 3, 4 if repetition is allowed.

Solution:

Given digits are 1, 2, 3, 4.

∴ Number of single digit numbers formed is 4.

Number of 2 digit numbers formed when repetitions is allowed is 4^{2} = 16.

Number of 3 – digit numbers formed when repetitions is allowed is 4^{3} = 64.

For 4 – digit number less than 2000, the thousands place is filled in 1 way, remaining 3 places can be filled in 4 ways each.

∴ Number of 4 digit numbers = 4^{3} = 64.

∴ Total number of numbers less than 2000 using given digits = 4 + 16 + 64 + 64 = 148.

III.

Question 1.

9 different letters of an alphabet are given. Find the number of 4 letter words that can be formed using these 9 letters which bave

i) no letter is repeated

ii) atleast one letter is repeated.

Solution:

The number of 4 letter words that can be formed using 9 different letters when repeti tion is allowed = 9^{4}

i) The number of 4 letter words that can be formed using 9 dIfferent letters when no letter is repeated = \({ }^9 \mathrm{P}_4\)

ii) The number of 4 letter words that can be formed using 9 different letters so that atleast one letter is repeated = 9^{4} – \({ }^9 \mathrm{P}_4\) = 3537.

Question 2.

Find the number of 4-digit numbers which can be formed using the digits 0, 2, 5, 7, 8 that are divisible by

(i) 2

(ii) 4 when repetition is allowed.

Solution:

Given digits are 0, 2, 5, 7, 8.

i) Divisible by 2:

The thousand’s place of 4 digit number when repetition is allowed can be filled in 4 ways. (using non-zero digits)

The 4-digit number is divisible by 2, when the units place is an even digit. This can be done in 3 ways.

The remaining 2 places can be filled by 5 ways each i.e., 5^{2} or 25 ways.

∴ Number of 4 digit numbers which are divisible by 2 is 4 × 3 × 25 = 300.

ii) Divisible by 4:

A number is divisible by 4 only when the number in last two places (tens and units) is a multiple of 4.

As repetition is allowed the last two places should be filled with one of the following 00, 08, 20, 28, 52, 72 80, 88

This can be done is 8 ways.

Thousands place is filled in 4 ways. (i.e., using non-zero digits)

Hundreds place can be filled in 5 ways.

∴ Total number of 4 digit numbers formed = 8 × 4 × 5 = 160.

Question 3.

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:

Given digits are 0, 1, 2, 3, 4, 5.

Thousands place can be filled in 5 ways, (using non-zero digit) when repetition is allowed.

Hundred’s place and ten’s place can be filled in 6 ways each, i.e., 62 ways.

If we fill up the unit’s place in 6 ways, we get 6 consecutive positive integers.

Out of any six consecutive integers only one is divisible by ‘6’.

Hence unit’s place is filled in 1 way.

Hence number of 4 digit numbers which are divisible by 6 using given digits = 5 × 6^{2} × 1 = 180.