Permutation questions deal with the arrangement of objects in a specific order or formation of a number of different words from the letters of a given word, etc. There exist a variety of cases in which we apply permutations to get the possible results. In this article, you will get solved questions on permutation, and some practice questions for the same.
What is Permutation?
Permutation refers to the arrangement of objects in a definite order. That means permutation is the arrangement of objects in which order matters. The arrangement of r objects out of n objects can be calculated using the permutation formula. That is:
nPr = n!/(n – r)!
Learn in detail about permutation here.
Permutation Questions and Answers
1. Calculate the following:
(i) nPr when n = 12, r = 5
(ii) 9P4
Solution:
(i) nPr when n = 12, r = 5
nPr = 12P5 = 12!/(12 – 5)!
= 12!/7!
= (12 × 11 × 10 × 9 × 8 × 7!)/7!
= 12 × 11 × 10 × 9 × 8
= 95040
(ii) 9P4
9P4 = 9!/(9 – 4)! = 9!/5!
= (9 × 8 × 7 × 6 × 5!)/5!
= 3024
2. In how many different ways can the letters of the word THOUGHTS be arranged so that the vowels always come together?
Solution:
Given word: THOUGHTS
Number of letters = 8
T’s = 2
H’s = 2
Number of vowels = 2 (O, U)
Vowels should come together.
So, the number of letters for arrangement = 7
i.e., (OU)THGHTS
Number of arrangements = 7!
And two vowels can be arranged in 2! ways.
Therefore, the total number of ways of arrangements = (7! × 2!)/(2! 2!)
= 7!/2!
= (7 × 6 × 5 × 4 × 3 × 2!)/2!
= 2520
3. In how many ways can seven books be arranged on a shelf?
Solution:
Number of ways in which the first book can be placed = 7
Number of ways in which the second book can be placed = 6
Similarly,
The total number of ways in which seven books can be arranged on a shelf = 7 × 6 × 5 × 4 × 3 × 2 × 1 (i.e., 7!)
= 5040
4. How many different arrangements of letters of the word MATHEMATICS are possible?
Solution:
Given word: MATHEMATICS
Number of letters = 11
M = 2
A = 2
T = 2
Number of different arrangements = 11!/(2! 2! 2!)
= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1 × 2 × 1 × 2 × 1)
= 4989600
5. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?
Solution:
Given digits: 1, 2, 3, 4, 6, 7
Number of digits = 6
Number of possible digits at unit’s place = 3 (2, 4 and 6)
⇒ Number of permutations = 3P1 = 3
When one of the digits is taken in units’ place, then the number of possible digits available = 5
⇒ Number of permutations = 5P2 = 5!/(5 – 2)! = 5!/3! = 120/6 = 20
The total number of permutations = 3 × 20 = 60.
Therefore, 60 three-digit numbers can be made using the given digits.
6. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Solution:
Given: 5 men and 4 women
Total number of people = 9
The women occupy even places, which means they will be sitting in the 2nd, 4th, 6th and 8th places, whereas the men will be sitting in the 1st, 3rd, 5th, 7th and 9th places.
The number of arrangements in which 4 women can sit in 4 places = 4P4 = 4!/(4 – 4)! = 4!/0! = 24/1 = 24
5 men can occupy 5 seats in 5 ways.
That means the number of ways they can be seated = 5P5 = 5!/(5 – 5)! = 5!/0! = 120/1 = 120
Therefore, the total numbers of possible sitting arrangements = 24 × 120 = 2880
7. How many numbers are there between 100 and 1000 such that at least one of their digits is 7?
Solution:
Total number of 3-digit numbers having at least one of their digits as 7 = (Total number of 3-digit numbers) – (Total number of 3-digit numbers in which 7 does not appear at all)
Let us find the total number of 3-digit numbers between 100 and 1000.
That means repetition of digits is allowed.
The hundred’s place can be filled in 9 ways, i.e., using digits from 1 to 9.
The ten’s and the unit’s place can be filled in 10 ways using the digits from 0 to 9.
∴ Total number of 3-digit numbers = 9 × 10 × 10 = 900
Now, we need to find the total number of 3-digit numbers in which 7 does not appear.
Here, 9 digits to be used, i.e., 0, 1, 2, 3, 4, 5, 6, 8, 9.
Now, the hundred’s place can be filled in 8 ways (excluding 0), and the tens’ and ones’ place can be filled in 9 ways each.
∴ Total number of 3-digit numbers in which 7 does not appear = 8 × 9 × 9 = 648
Hence, the numbers between 100 and 1000 in which at least one of their digits is 7 = 900 – 648 = 252
8. P, Q, R, S, and T sit on five chairs facing north. R will sit only on the leftmost chair, and Q will not sit anywhere to the left of P. In how many ways can they be seated?
Solution:
R will sit on 1 and Q will sit somewhere to the right of P
1 2 3 4 5
R will sit on 1
Then there are three possible ways.
Case 1:
P on 2, so Q can be seated on 3, 4 or 5
The remaining two can be seated on two chairs in 2 ways
Number of possible ways = 3 × 2 = 6
Case 2:
P on 3, so Q can be seated on 4 or 5
Number of possible ways = 2 × 2 = 4
Case 3:
P on 4, so Q will be on 5
Number of possible ways = 2
Total number of possible ways = (6 + 4 + 2) = 12
9. In how many ways can 10 differently coloured beads be threaded on a string?
Solution:
As the necklace can be turned over, clockwise and anti-clockwise arrangements are the same.
Also, the number of string arrangements of n objects = (n – 1)!/2
Therefore, the number of ways in which 10 differently coloured beads can be threaded on a string = (10 – 1)!/2 = 9!/2
= 181440
10. How many 3 digit numbers can be formed with the digits 5, 6, 2, 3, 7 and 9, which are divisible by 5, and none of its digits are repeated?
Solution:
If the number has unit’s digits as 0 or 5, then it will be divisible by 5.
Given digits: 5, 6, 2, 3, 7, 9
Number of digits = 6
The digit that can be placed at the unit’s place so that the three-digit number is divisible by 5 is 5.
This can be done in 1 way.
Now, the tens and hundreds of places can be filled in 5 and 4 ways since the repetition of the digits are not allowed.
Therefore, the total number of such three digits numbers = 5 × 4 × 1 = 20
Practice Problems on Permutation
- How many ways can the letters of the word “EXAMINATION” be arranged such that the first and last letters are the same, and the vowels are together?
- How many numbers between 100 and 1000 use only odd digits, no digit being repeated?
- How many 3-letter words, with or without meaning, can be formed out of the letters of the word LOGARITHMS, if repetition of letters is not allowed?
- In how many ways can 9 different colour balls be arranged in a row so that the black, white, red and green balls are never together?
- Find the sum of all the possible numbers of 4 digits formed by digits 3, 5, 5, and 6 using each digit once.
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