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RD Chapter 16- Permutations Ex-16.4 Interview Questions Answers

Question 1 : In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?

Answer 1 :

Given:

The word ‘FAILURE’

Number of vowels inword ‘FAILURE’ = 4(E, A, I, U)

Number of consonants =3(F, L, R)

Let consonants bedenoted by C

Odd positions are 1,3, 5 or 7

The consonants can bearranged in these 4 odd places in 4P3 ways.

Remaining 3 evenplaces (2, 4, 6) are to be occupied by the 4 vowels. This can be done in 4P3 ways.

So, the total numberof words in which consonants occupy odd places = 4P3 × 4P3

By using the formula,

P (n, r) = n!/(n – r)!

P (4, 3) × P (4, 3) =4!/(4 – 3)! × 4!/(4 – 3)!

= 4 × 3 × 2 × 1 × 4 ×3 × 2 × 1

= 24 × 24

= 576

Hence, the number ofarrangements so that the consonants occupy only odd positions is 576.

Question 2 :
In how many ways can the letters of the word ‘STRANGE’ be arranged so that
(i) the vowels come together?
(ii) the vowels never come together? And
(iii) the vowels occupy only the odd places?

Answer 2 :

Given:

The word ‘STRANGE’

There are 7 letters inthe word ‘STRANGE’, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G).

(i) the vowels cometogether?

Considering 2 vowelsas one letter so we will have 6 letters which can be arranged in 6P6 ways.

(A,E) can be puttogether in 2P2 ways.

Hence, the requirednumber of words are

By using the formula,

P (n, r) = n!/(n – r)!

P (6, 6) × P (2, 2) =6!/(6 – 6)! × 2!/(2 – 2)!

= 6! × 2!

= 6 × 5 × 4 × 3 × 2 ×1 × 2 × 1

= 720 × 2

= 1440

Hence, total number ofarrangements in which vowels come together is 1440.

(ii) the vowels never cometogether?

The total number ofletters in the word ‘STRANGE’ is 7P7 = 7! = 7 ×6 × 5 × 4 × 3 × 2 × 1 = 5040

So,

Total number of wordsin which vowels never come together = total number of words – number of wordsin which vowels are always together

= 5040 – 1440

= 3600

Hence, the totalnumber of arrangements in which vowel never come together is 3600.

(iii) the vowels occupy onlythe odd places?

There are 7 letters inthe word ‘STRANGE’. Out of these letters (A,E) are the vowels.

There are 4 odd placesin the word ‘STRANGE’. The two vowels can be arranged in 4P2 ways.

The remaining 5consonants an be arranged among themselves in 5P5 ways.

So, the total numberof arrangements is

By using the formula,

P (n, r) = n!/(n – r)!

P (4, 2) × P (5, 5) =4!/(4 – 2)! × 5!/(5 – 5)!

= 4!/2! × 5!

= (4×3×2!)/2! × 5!

= 4×3 × 5 × 4 × 3 × 2× 1

= 12 × 120

= 1440

Hence, the number ofarrangements so that the vowels occupy only odd positions is 1440.

Question 3 : How many words can be formed from the letters of the word ‘SUNDAY’? How many of these begin with D?

Answer 3 :

Given:

The word ‘SUNDAY’

Total number ofletters in the word ‘SUNDAY’ is 6.

So, number ofarrangements of 6 things, taken all at a time is 6P6

= 6! = 6 × 5 × 4 × 3 ×2 × 1 = 720

Now, we shall find thenumber of words starting with D

So let’s fix the firstposition with letter D, then remaining number of letters is 5.

The number ofarrangements of 5 things, taken all at a time is 5P5 =5! = 5 × 4 × 3 × 2 × 1 = 120

Hence, the totalnumber of words can be made by letters of the word ‘SUNDAY’ is 720.

The possible number ofwords using letters of ‘SUNDAY’ starting with ‘D’ is 120.

Question 4 : How many words can be formed out of the letters of the word, ‘ORIENTAL,’ so that the vowels always occupy the odd places?

Answer 4 :

Given:

The word ‘ORIENTAL’

Number of vowels inthe word ‘ORIENTAL’ = 4(O, I, E, A)

Number of consonantsin given word = 4(R, N, T, L)

Odd positions are (1,3, 5 or 7)

Four vowels can bearranged in these 4 odd places in 4P4 ways.

Remaining 4 evenplaces (2,4,6,8) are to be occupied by the 4 consonants in 4P4 ways.

So, by using theformula,

P (n, r) = n!/(n – r)!

P (4, 4) × P (4, 4) =4!/(4 – 4)! × 4!/(4 – 4)!

= 4! × 4!

= 4 × 3 × 2 × 1 × 4 ×3 × 2 × 1

= 24 × 24

= 576

Hence, the number ofarrangements so that the vowels occupy only odd positions is 576.

Question 5 : How many different words can be formed with the letters of word ‘SUNDAY’? How many of the words begin with N? How many begin with N and end in Y?

Answer 5 :

Given:

The word ‘SUNDAY’

Total number ofletters in the word ‘SUNDAY’ is 6.

So, number ofarrangements of 6 things, taken all at a time is 6P6

= 6! = 6 × 5 × 4 × 3 ×2 × 1 = 720

Now, we shall find thenumber of words starting with N

So let’s fix the firstposition with letter N, then remaining number of letters is 5.

The number ofarrangements of 5 things, taken all at a time is 5P5 =5! = 5 × 4 × 3 × 2 × 1 = 120

Now, we need to findout a number of words starting with N and ending with Y

So let’s fix the firstposition with letter N and Y at the end, then remaining number of letters is 4which can be arranged in 4P4 ways. = 4! = 4 × 3× 2 × 1 = 24

Hence, the totalnumber of words can be made by letters of the word ‘SUNDAY’ is 720.

The possible number ofwords using letters of ‘SUNDAY’ starting with ‘N’ is 120.

The possible number ofwords using letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ is 24.

Question 6 :
How many different words can be formed from the letters of the word ‘GANESHPURI’? In how many of these words:
(i) the letter G always occupies the first place?
(ii) the letter P and I respectively occupy the first and last place?
(iii) Are the vowels always together?
(iv) the vowels always occupy even places?

Answer 6 :

Given:

The word ‘GANESHPURI’

There are 10 lettersin the word ‘GANESHPURI’. The total number of words formed is 10P10 =10!

(i) the letter G alwaysoccupies the first place?

If we fix the firstposition with letter G, then remaining number of letters is 9.

The number ofarrangements of 9 things, taken all at a time is 9P9 =9! Ways.

Hence, a possiblenumber of words using letters of ‘GANESHPURI’ starting with ‘G’ is 9!

(ii) the letter P and Irespectively occupy the first and last place?

If we fix the firstposition with letter P and I in the end, then remaining number of letters is 8.

The number ofarrangements of 8 things, taken all at a time is 8P8 =8! Ways.

Hence, a possiblenumber of words using letters of ‘GANESHPURI’ starting with ‘P’ and ending with‘I’ is 8!

(iii) Are thevowels always together?

There are 4 vowels and6 consonants in the word ‘GANESHPURI’.

Consider 4 (A,E,I,U)vowels as one letter, then total number of letters is 7 (A,E,I,U, G, N, S, H ,P, R)

The number ofarrangements of 7 things, taken all at a time is 7P7 =7! Ways.

(A, E, I, U) can beput together in 4! Ways.

Hence, total number ofarrangements in which vowels come together is 7! × 4!

(iv) the vowels alwaysoccupy even places?

Number of vowels inthe word ‘GANESHPURI’ = 4(A, E, I, U)

Number of consonants =6(G, N, S, H, R, I)

Even positions are 2,4, 6, 8 or 10

Now, we have toarrange 10 letters in a row such that vowels occupy even places. There are 5even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even placesin 5P4 ways.

Remaining 5 odd places(1, 3, 5, 7, 9) are to be occupied by the 6 consonants in 6P5 ways.

So, by using theformula,

P (n, r) = n!/(n – r)!

P (5, 4) × P (6, 5) =5!/(5 – 4)! × 6!/(6 – 5)!

= 5! × 6!

Hence, number ofarrangements so that the vowels occupy only even positions is 5! × 6!


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RD Chapter 16- Permutations Ex-16.4 Contributors

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