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

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

I.

Question 1.

Find the number of ways of arranging the letters of the words.

i) INDEPENDENCE

ii) MATHEMATICS

iii) SINGING

iv) PERMUTATION

v) COMBINATION vQ INTERMEDIATE

Solution:

i) Given word is INDEPENDENCE

The given 12 letters word INDEPENDENCE contains 3N’s, 2D’s and 4I’s.

Hence they can be arranged in \(\frac{12 !}{2 ! \times 3 ! \times 4 !}\) ways.

ii) Given word is MATHEMATICS

The given 11 letters word MATHEMATICS contains 2 M’s, 2 A’s and 2 T’s.

Hence they can be arranged in \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

iii) Given word is SINGING.

The given 7 letters word SINGING contains 2 I’s, 2 N’s and 2 G’s.

Hence they can be arranged in \(\frac{7 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

iv) Given Word is PERMUTATION

The given 11 letters word PERMUTATION contains 2T’s.

Hence they can be arranged in \(\frac{11 !}{2 !}\) ways.

v) Given word is COMBINATION.

Given 12 letters word COMBINATION contains 2 O’s, 2 I’s, 3 N’s.

Hence they can be arranged in \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

vi) Given word is INTERMEDIATE.

Given 12 letters word INTERMEDIATE’ contains 2 I s, 2 T’s, 3 E’s.

Hence they can be arranged in \(\frac{12 !}{3 ! \times 2 ! \times 2 !}\) ways.

Question 2.

Find the number of 7- digit numbers that can be formed using 2, 2, 2, 3, 3, 4, 4.

Solution:

The 7 digit numbers that can be formed using three 2’s, two 3’s and two 4’s are \(\frac{7 !}{3 ! \cdot 2 ! \cdot 2 !}\).

II.

Question 1.

Find the number of 4 – letter words that can be formed using the letters of the word RAMANA.

Solution:

Given word is RAMANA.

The given word RAMANA has 6 letters which contains 3 A’s and rest are different.

In forming 4 letter words, these cases arises.

Case – (i) :

When all are different i.e., R, M, N, A.

∴ Number of 4 letter words formed = 4! = 24.

Case – (ii) :

When two are alike (i.e., A, A) and two are different, i.e., selected from R, M, N.

Two different letters are selected in \({ }^3 \mathrm{C}_2\) ways.

∴ Number of 4 letter words formed are \({ }^3 C_2 \times \frac{4 !}{2 !}\) = 36.

Case (iii):

When three are alike (i.e., A, A, A) and one different letter, selected out of R, M, N.

Now one different letters is selected in \({ }^3 \mathrm{C}_1\) ways.

∴ Number of 4 letter words formed = \({ }^3 \mathrm{C}_1 \times \frac{4 !}{3 !}\) = 12

∴ Total number of 4 letter words formed are 24 + 36 + 12 = 72.

Question 2.

How many numbers can be formed using all digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places?

Solution:

Given digits 1, 2, 3, 4, 3, 2, 1 contains 3 even digits i.e., 2, 4, 2 and 3 even places.

Number of ways of arranging 3 even digits 2, 4, 2 in three even places is \(\frac{3 !}{2 !}\) 3.

The remaining 4 digits 1, 3, 3, 1 in remaining 4 places can be arranged in \(\frac{4 !}{2 ! \cdot 2 !}\) ways.

∴ Number of numbers formed using all the digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places = 3 × \(\frac{4 !}{2 ! \cdot 2 !}\) = 18.

Question 3.

In a library, there are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies each of two different books. Find the number of ways of arranging all these books in a shelf in a single row.

Solution:

There are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies of two different books.

∴ Total number of books = 6 + 4(2) + 5(3) + 3(2) = 35

∴ Number of ways of arranging these 35 books = \(\frac{35 !}{6 ! \times 4 ! \times 4 ! \times 5 ! \times 5 ! \times 5 ! \times 3 ! \times 3 !}\)

= \(\frac{35 !}{6 ! \cdot(4 !)^2 \cdot(5 !)^3 \cdot(3 !)^2}\).

Question 4.

A book store has ‘m’ copies each of ‘n’ different books. Find the number of ways of arranging these books in a shelf in a single row.

Solution:

Book store has m’ copies each of n’ different books.

∴ Total number of books = mn

∴ Number of ways of arranging these mn books = \(\frac{(m n) !}{m ! \times m ! \times \underbrace{\ldots \ldots m !}_{n \text { times }}}=\frac{(m n) !}{(m !)^n}\)

Question 5.

Find the number of 5-digit numbers that can be formed using the digits 0, 1, 1, 2, 3.

Solution:

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

The ten thousand’s place of a ‘5’ digit number can be filled by any of non-zero digit in 4 ways.

The remaining 4 places can be filled with remaining 4 digits in 4! ways.

∴ Number of ways = 4 × 4!

Since there are two 1’s in every arrangement, the number of 5 digited numbers formed are \(\frac{4 \times 4 !}{2 !}\) = 48.

Question 6.

In how many ways can the letters of the word CHEESE be arranged so that no two E’s come together ?

Solution:

Given word is ‘CHEESE’.

As no two E’s come together, first arrange letters C, H, S.

Number of ways of arranging C, H, S is 3!.

∴ Number of gaps formed are 4.

Number of ways of arranging 3 E’s in 4 gaps = \(\frac{{ }^4 P_3}{3 !}\)

∴ Required number of ways = 3! × \(\frac{{ }^4 P_3}{3 !}\) = 24.

III.

Question 1.

Find the number of ways of arranging the letters of the word ASSOCIATIONS. In how many of them

i) all the three S’s together

ii) the two A’s do not come together.

Solution:

Given word is ‘ASSOCIATIONS’.

The given 12 letters word ‘ASSOCIATIONS’ contains 2 A’s, 3 S’s, 2 O’s and 2 I s.

∴ Number of ways of arranging them = \(\frac{12 !}{2 ! \cdot 2 ! \cdot 2 ! \cdot 3 !}\).

i) All the three S’s come together :

Treat 3 S’s as one unit. This unit with remain¬ing 9 letters becomes 10 entities which contain 2 A’s, 2 O’s and 2 I’s.

Number of ways of arranging so that all the three S’s come together = \(\frac{10 !}{2 ! \cdot 2 ! \cdot 2 !}\).

ii) The two A’s do not come together :

As 2 A’s do not come together, arrange remaining 10 letters.

Remaining 10 letters contains 3 S’s, 2 O’s and 2 I’s.

∴ Number of ways of arranging = \(\frac{10 !}{3 ! \cdot 2 ! \cdot 2 !}\)

∴ Number of gaps formed are ’11’.

2 A’s in these 11 gaps can be arranged in up \(\frac{{ }^{11} \mathrm{P}_2}{2 !}\) ways.

∴ Required number of ways of arranging = \(\frac{10 !}{3 ! \times 2 ! \times 2 !} \times \frac{{ }^{11} P_2}{2 !}\).

Question 2.

Find the number of ways of arranging the letters of the word MISSING so that the two S’s are together and the two l’s are together.

Solution:

Given word is MISSING.

Treat 2 S’s as 1 unit and 2 l’s as another unit.

These 2 unit with remaining 3 letters be comes 5 entities.

Number of ways of arranging them = 5!

2 S’s and 21’s can arrange themselves in only 1 way.

∴ Required nunther of ways of arranging = 5! = 120.

Question 3.

If the letters of the word AJANTA are per muted in all possible ways and the words thus formed are arranged In dictionary order, find the ranks of the words

(i) AJANTA

(ii) JANATA.

Solution:

Given word MANTA in Alphabetic order is AAAJNT.

i) Rank of ‘AJANTA’:

In the dictionary order first comes that words which begin with letter A. Also the second place is to be filled by A.

The remaining 4 places can be filled in 4! ways.

On proceeding like this, we get

The number of word begin with AA is 4! = 24

The number of word begin with AJAA is 2! = 2

The number of word begin with AJANA = 1

Next word is AJANTA

∴ Rank of word AJANTA = 24 + 2 + 1 + 1 = 28.

ii) Rank of JANATA :

In the dictionary order first comes that words which begin with the letter A.

The remaining 5 places to be filled in \(\frac{5 !}{2 !}\) ways.

On proceeding like this

The number of words begin with A = \(\frac{5 !}{2 !}\) = 60.

The number of words begin with JAA = 3! = 6

The number of words begin with JANAA = 1! = 1

Next word is JANATA

∴ Rank of word JANATA = 60 + 6 + 1 + 1 = 68.