In this section, we shall discuss the permutations where either the repetition of items is allowed or the distinction between some of the items are ignored, or a particular item occurs in every arrangement etc. Such types of permutations are termed permutations under certain conditions. Subject experts have prepared the solutions to improve the confidence of students to appear for the board exam without fear. The solutions PDFs are developed based on the requirement of students, which reduces time constraints and helps students solve problems with ease. In order to excel in the board exams with high marks, RD Sharma Class 11 Maths Solutions PDF is available here for free download.
RD Sharma Solutions for Class 11 Maths Exercise 16.4 Chapter 16 – Permutations
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1. In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?
Solution:
Given:
The word ‘FAILURE’
Number of vowels in word ‘FAILURE’ = 4(E, A, I, U)
Number of consonants = 3(F, L, R)
Let consonants be denoted by C
Odd positions are 1, 3, 5 or 7
The consonants can be arranged in these 4 odd places in 4P3 ways.
The remaining 3 even places (2, 4, 6) are to be occupied by the 4 vowels. This can be done in 4P3 ways.
So, the total number of 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 of arrangements so that the consonants occupy only odd positions is 576.
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?
Solution:
Given:
The word ‘STRANGE’
There are 7 letters in the word ‘STRANGE’, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G).
(i) The vowels come together.
Considering 2 vowels as one letter so we will have 6 letters which can be arranged in 6P6 ways.
(A,E) can be put together in 2P2 ways.
Hence, the required number 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 of arrangements in which vowels come together is 1440.
(ii) The vowels never come together.
The total number of letters in the word ‘STRANGE’ is 7P7 = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
So,
Total number of words in which vowels never come together = total number of words – number of words in which vowels are always together
= 5040 – 1440
= 3600
Hence, the total number of arrangements in which vowels never come together is 3600.
(iii) The vowels occupy only the odd places.
There are 7 letters in the word ‘STRANGE’. Out of these letters (A,E) are the vowels.
There are 4 odd places in the word ‘STRANGE’. The two vowels can be arranged in 4P2 ways.
The remaining 5 consonants can be arranged among themselves in 5P5 ways.
So, the total number of 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 of arrangements so that the vowels occupy only odd positions is 1440.
3. How many words can be formed from the letters of the word ‘SUNDAY’? How many of these begin with D?
Solution:
Given:
The word ‘SUNDAY’
The total number of letters in the word ‘SUNDAY’ is 6.
So, the number of arrangements of 6 things, taken all at a time, is 6P6
= 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Now, we shall find the number of words starting with D
So let’s fix the first position with the letter D, and then the remaining number of letters is 5.
The number of arrangements of 5 things, taken all at a time, is 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120
Hence, the total number of words that can be made by letters of the word ‘SUNDAY’ is 720.
The possible number of words using letters of ‘SUNDAY’ starting with ‘D’ is 120.
4. How many words can be formed out of the letters of the word, ‘ORIENTAL,’ so that the vowels always occupy the odd places?
Solution:
Given:
The word ‘ORIENTAL’
Number of vowels in the word ‘ORIENTAL’ = 4(O, I, E, A)
Number of consonants in given word = 4(R, N, T, L)
Odd positions are (1, 3, 5 or 7)
Four vowels can be arranged in these 4 odd places in 4P4 ways.
The remaining 4 even places (2,4,6,8) are to be occupied by the 4 consonants in 4P4 ways.
So, by using the formula,
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 of arrangements so that the vowels occupy only odd positions is 576.
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?
Solution:
Given:
The word ‘SUNDAY’
The total number of letters in the word ‘SUNDAY’ is 6.
So, the number of arrangements of 6 things, taken all at a time, is 6P6
= 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Now, we shall find the number of words starting with N
So let’s fix the first position with the letter N, and then remaining number of letters is 5.
The number of arrangements of 5 things, taken all at a time, is 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120
Now, we need to find out a number of words starting with N and ending with Y
So let’s fix the first position with letters N and Y at the end, then the remaining number of letters is 4, which can be arranged in 4P4 ways. = 4! = 4 × 3 × 2 × 1 = 24
Hence, the total number of words that can be made by letters of the word ‘SUNDAY’ is 720.
The possible number of words using the letters of ‘SUNDAY’ starting with ‘N’ is 120.
The possible number of words using the letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ is 24.
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?
Solution:
Given:
The word ‘GANESHPURI’
There are 10 letters in the word ‘GANESHPURI’. The total number of words formed is 10P10 = 10!
(i) The letter G always occupies the first place.
If we fix the first position with letter G, then remaining number of letters is 9.
The number of arrangements of 9 things, taken all at a time is 9P9 = 9! Ways.
Hence, a possible number of words using letters of ‘GANESHPURI’ starting with ‘G’ is 9!
(ii) The letters P and I respectively occupy the first and last place.
If we fix the first position with letter P and I in the end, then remaining number of letters is 8.
The number of arrangements of 8 things, taken all at a time is 8P8 = 8! Ways.
Hence, a possible number of words using letters of ‘GANESHPURI’ starting with ‘P’ and ending with ‘I’ is 8!
(iii) The vowels are always together?
There are 4 vowels and 6 consonants in the word ‘GANESHPURI’.
Consider 4 (A,E,I,U) vowels as one letter, then the total number of letters is 7 (A,E,I,U, G, N, S, H , P, R)
The number of arrangements of 7 things, taken all at a time, is 7P7 = 7! Ways.
(A, E, I, U) can be put together in 4! Ways.
Hence, the total number of arrangements in which vowels come together is 7! × 4!
(iv) The vowels always occupy even places.
Number of vowels in the 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 to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even places in 5P4 ways.
The remaining 5 odd places (1, 3, 5, 7, 9) are to be occupied by the 6 consonants in 6P5 ways.
So, by using the formula,
P (n, r) = n!/(n – r)!
P (5, 4) × P (6, 5) = 5!/(5 – 4)! × 6!/(6 – 5)!
= 5! × 6!
Hence, the number of arrangements so that the vowels occupy only even positions is 5! × 6!
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