The aim of this topic is to help you build your basics in Permutation and Combination. In the section introduction to Permutation and Combination we have shared some important terminologies involved in Permutation and Combination problems. The exercises contain solved question and answers. The key to solving a problem is knowing how to approach the problem. When solving exercises, we recommend you to try solving the problems on your own, then view the solutions. That way we believe you can strengthen your approach in solving problems on Permutations and Combinations. By the end of this topic, you will learn about:
- Fundamental counting principle
- Permutation and Combination and their relation
- Permutation when some are identical
- Circular permutations
Introduction to Permutation and Combination
Fundamental Principle of Counting
Product Rule
If there are m ways of doing one thing and n ways of doing another, then there are m x n ways of doing both.
Example : If there are 10 shirts and 12 trousers, then the total number of choices of a unique outfit is 10 x 12 = 120
Addition Rule
If there are m ways of doing one thing and n ways of doing another, then there are m + n ways of doing either.
Example : If there are m ways of doing one thing and n ways of doing another, then there are m + n ways of choosing one garment.
Factorial
Factorial of n (represented as n!) = n x (n-1) x (n-2) x …1
Permutation
It is the arrangement of r things from n dissimilar things at a time without replacement. It is denoted by ^nP_r
The arrangement of r persons from n persons is given by
$^nP_r = \frac{n!}{(n-r)!}$
Example:
The number of ways in which 2 persons can be arranged from set of 3 persons is
$^3P_2 = \frac{3!}{(3-2)!}$
= 6
nP0 =1, nP1 = n, nPn = n!
Combination
It is the selection of r things at a time from n things without replacement. It is denoted by n C r and given by
nCr = $\frac{n!}{(n-r)!r!}$
Example:
The number of ways of selecting 2 persons from a group of 3 persons is
3C2 = $\frac{3!}{(3-2)!2!}$
= 3
nC0 = nCn =1 nC1 = nCn-1 = n nCr = nCn-r
Relation between Permutation and Combination
Combination => Selection of r persons
Permutation = > Selection of r persons x Arrangement of r persons
nPr = nCr x r!
Permutation of things when some are identical.
If there are n things out of which p things are alike of one kind, q things are alike of another kind and r things are alike of the third and the rest are all different, then the total number of arrangements of n things taken all at a time is
$\frac {n!}{p! q! r!}$
Circular Permutation
The number of circular permutations of n things taken all at a time is given by (n-1)!
In case of a garland or a necklace, the number of circular permutations is $\frac{(n-1)!}{2}$, since the arrangement is same both clockwise and anticlockwise.
Note:
- 0! = 1! = 1
Problems
Permutation and Combination - Exercise - Blog
20 questions
- There are 2 routes from A to B and 1 route from A to C. B to D has 3 routes, while C to D has 2 routes, Find the number of ways in which a person can go from A to D.
- 2
- 6
- 8
- 10
Show the answer
Answer: 8
A person can reach D either via B or via C.
Via B :
Since there are 2 routes from A to B and 3 routes from B to D, using the product rule of counting principle A to D via B can be reached in 2 x 3 = 6 ways.
Via C :
Since there is 1 route from A to C and 2 routes from C to D, using the product rule of counting principle A to D via C can be reached in 1 x 2 = 2 ways.
Since a person can reach D either through B or through C, using Addition rule of counting principle, total number of ways = 6 + 2 = 8
- Find the number of one to one functions from set A of 3 elements to set B of 4 elements.
- 7
- 9
- 12
- 24
Show the answer
Answer: 24
For the first element in set A there are 4 ways of choosing an element in set B. Similarly second element in set A has 3 choices in set B and third element in set A has 2 choices in set B.
Hence, the number of one-one functions = 4 x 3 x 2 = 24.
- Find the number of arrangements of the letters of the word PRODUCT, so that the vowels always come together.
- 720
- 1440
- 20
- 40
Show the answer
Answer: 1440
There are 2 vowels O and U in PRODUCT.
Since vowels need to be together, consider O and U as one letter say X. We thus need to find the arrangement of 6 letters in PRDCTX.
The 6 letters can be arranged in 6! ways (First letter can be placed in 6 positions, second letter in 5 positions and so on = 6 x 5 x 4 x 3 x 2 x 1)
The two vowels O and U can be arranged in 2! ways.
Therefore, the total number of ways = 6! x 2! = 1440
- Find the number of 4 digit even numbers that can be formed, if repetition is allowed.
- 4500
- 5000
- 9000
- 10000
Show the answer
Answer: 4500
For a number to be even, the units digit must be even. Hence from the ten digits - 0,1,2,3,4,5,6,7,8,9
- The units digit can be filled in 5 ways (0 or 2 or 4 or 6 or 8)
- The digit in the thousands place can be filled in 9 ways (Since 0 cannot be in thousand’s place)
- The digit in hundreds place can be filled in 10 ways (digits from 0 to 9)
- The digit in tens place can be filled in 10 ways (digits from 0 to 9)
Hence total number of 4 digit even numbers = 9 x 10 x 10 x 5 = 4500
- Find the number of 3 digit numbers that can be formed using the digits 2, 5, 6 and 7 if repetition is not allowed.
- 64
- 24
- 60
- 42
Show the answer
Answer: 24
From 2,5,6 and 7
- Hundreds place can filled in 4 ways.
- Tens place can be filled in 3 ways (Since one digit is already placed in hundreds place and repetition is not allowed)
- Similarly, units place can be filled in 2 ways
Hence total number of 3 digit numbers using 2, 5, 6 and 7 is 4 x 3 x 2 = 24
- Find the number of ways in which a group of 4 boys and 4 girls can be seated in a straight line such that no 2 boys sit together.
- 24
- 120
- 1440
- 2880
Show the answer
Answer: 2880
Since no 2 boys sit together, the possible seating arrangement in a straight line is
_G_G_G_G_
From the arrangement,
- 4 girls can be arranged in 4 positions in 4P4 ways = $\frac{4!}{(4-4)!}$ = 4x3x2x1 = 24
- There are 5 possible positions where the 4 boys can be arranged, which can be done in 5P4 ways = $\frac{5!}{(5-4)!}$ = 5 x 4 x 3 x 2 x 1 = 120
Total number of arrangements = 120 x 24 = 2880
- Number of 4 letter words that can be formed from the letters of the word PROBLEM if repetition is not allowed.
- 4
- 6
- 12
- 15
Show the answer
Answer: 15
Forming 4 letters from 6 letters of the word PROBLEM, is an arrangement of 4 letters. Number of such 4 letter word is obtained by using the permutation :
6P4 = $\frac{6!}{{(6-4)!}}$
= 6 x 5 / 2 = 15
- There are 3 different types of buses, 4 different types of cars and 5 different types of bikes. Find the number of ways in which the vehicles can be arranged such that all buses are together, all cars are together and all bikes are together.
- 103680
- 3620
- 2880
- 102460
Show the answer
Answer: 103680
Let all the buses be denoted as X, cars be Y and bikes be Z.
Then the number of ways in which the vehicles can be arranged can be calculated by taking the product of values obtained from the following four steps:
- Arrangement of X,Y and Z = 3! = 6
- Arrangement of all the different types of buses = 3! = 6
- Arrangement of all the different types of cars = 4! = 24
- Arrangement of all the different types of bikes = 5! = 120
Total number of ways = 6 x 6 x 24 x120 = 103680
- Find the number of arrangements of the letters of the word QUANT so that the vowels are in odd places.
- 12
- 24
- 30
- 36
Show the answer
Answer: 36
There are 2 vowels and 3 odd places. Hence, the 2 vowels can be placed in 3 odd places in 3P2 = 6 ways.
The remaining 3 characters can be placed in the remaining 3 places in 3P3 = 6 ways.
From the fundamental counting principle, total number of ways = 6 x 6 = 36 ways.
- How many numbers between 100 and 10000 can be formed using the digits 2, 3, 5, 7 and 9 without repetition.
- 120
- 60
- 180
- 40
Show the answer
Answer: 180
Case 1 : Number of 3 digit numbers between 100 and 10000 using the digits 2, 3, 5, 7 and 9
- Hundreds place can be filled in 5 ways
- Tens place can be filled in 4 ways
- Units place can be filled in 3 ways
Total number of 3 digit numbers based on the given condition = 5 x 4 x 3 = 60
Case 2 : Number of 4 digit numbers between 100 and 10000 using the digits 2, 3, 5, 7 and 9
- Thousands place can be filled in 5 ways
- Hundreds place can be filled in 4 ways
- Tens place can be filled in 3 ways
- Units place can be filled in 2 ways
Total number of 4 digit numbers based on the given condition = 5 x 4 x 3 x 2 = 120
The total numbers between 100 and 10000 formed using the digits 2, 3, 5, 7 and 9 without repetition is the sum of the values obtained in Case 1 and Case 2 = 60 + 120 = 180
- A bag has 5 red balls, 4 black balls and 6 white balls. Find the number of ways in which 3 red balls, 4 white balls and 2 black balls can be chosen from the bag.
- 300
- 450
- 700
- 900
Show the answer
Answer: 900
3 red balls can be chosen in 5C3 = 10 ways
4 white balls can be chosen in 6C4 = 15 ways
2 black balls can be chosen in 4C2 = 6 ways
Total number of ways = 10 x 15 x 6 = 900
- From a group of 5 men and 4 women a committee of 4 members had to be formed. In how many ways can this committee be formed, if at least 1 man and 1 woman must be there in the committee.
- 20
- 40
- 60
- 80
Show the answer
Answer: 60
Case 1 : 1 Man and 3 Women
1 man and 3 Women can be selected in 5C1 x 4C3 = 20 ways
Case 2 : 2 Men and 2 Women
2 Men and 2 Women can be selected in 5C2 x 4C2 = 60 ways
Case 3 : 3 Men and 1 Woman
3 Men and 1 Woman can be selected in 5C3 x 4C1 ways = 60 ways
- In an exam, there are 100 multiple choice question each with 4 options. How many answer keys are possible?
- 100
- 400
- $4^{100}$
- ${100}^4$
Show the answer
Answer: $4^{100}$
For each question there are 4 answers are possible. Therefore for 100 questions, the number of possible answer keys = 4 x 4 x 4… 100 questions = $4^{100}$
- In a party, there are 30 persons. Each person shakes hand with every other person. Find the number of handshakes.
- 15
- 30
- 300
- 435
Show the answer
Answer: 435
Every time an handshake takes place 2 out of 30 members are involved. The total number of handshakes is obtained by finding the total number of ways of selecting 2 people from 30, which is given 30C2 = 435
- How many diagonals can be formed by joining the vertices of a hexagon.
- 15
- 6
- 9
- 30
Show the answer
Answer: 9
A diagonal is formed when we join 2 points by a line. An hexagon has 6 vertices. Hence, the total number of lines in a regular hexagon = 6C2 = 15.
Out of the 15 lines, 6 lines form the sides of the hexagon. The remaining lines are the diagonals, which is 9.
- In how many ways can the letters of the word JUSTQUANT be arranged so that the vowels always appear together.
- 2520
- 5040
- 7560
- 10080
Show the answer
Answer: 7560
JUSTQUANT has 2 U’s, 2 T’s.
Let the vowels A and 2 U’s be considered as single character X.
- Then, the total number of arrangements of remaining 7 characters is $\frac{7!}{2!}$ (2! because of 2 T’s) = 2520
- A and 2 U’s can be arranged in $\frac{3!}{2!}$ = 3 ways.
Total number of ways = 2520 x 3 = 7560
- In how many ways can the letters of the word BRILLIANT be arranged so that the vowels are in odd places.
- 180
- 360
- 900
- 1800
Show the answer
Answer: 1800
There are 3 vowels - 1 A and 2 I’s, and 5 odd places. Hence these 3 vowels can be placed in 5 odd places in $\frac{^5P_3}{2!}$ = 5 ways
The remaining 6 characters - B, R, 2 L’s, N, T can be placed in remaining 6 places in $\frac{^6P_6}{2!}$ = 360.
The total number of ways = 360 x 5 = 1800.
- How many 4 letter words can be formed from the letters of the word COMMITTEE.
- 36
- 180
- 360
- 576
Show the answer
Answer: 576
There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's, 1C, 1O, 1I. Hence there are totally 6 different letters in the letters of the word COMMITTEE.
Case 1 : All 4 letter are different - abcd
We can choose 4 letters from 6 letters in 6C4 ways and these 4 can be arrange in 4! ways. Hence the total number of ways => 6C4 x 4! = 360
Case 2 : 2 letters same and 2 different - aabc
We can choose repeated letters in 3C1 ways and 2 distinct letters from the remaining 5 categories in 5C2 ways and we arrange them in $$\frac{4!}{2!}$ ways. Hence total ways => 180
Case 3 : 2 letters repeat twice - aabb
We choose the repeated letter from 4 categories in 4C2 ways and we arrange them in $\frac{4!}{2!x2!}$ ways. So total ways 4C2 x $\frac{4!}{2!x2!}$ = 36
4 letter words that can be formed from the letters of the word COMMITTEE can be obtained by adding all the cases :
360 + 180 + 36 = 576
- Find the number of ways in which 4 boys and 4 girls can be seated in a round table, so that no 2 girls sit together.
- 24
- 64
- 128
- 144
Show the answer
Answer: 144
The 4 boys are seated round the table in 4 chairs such that a girl can sit between every 2 boys. This is possible (4-1)! = 6 ways.
For each arrangement of the boys the girls can sit in 4! ways = 24 ways.
Therefore, total number of ways = 24x6 = 144.
- Find the number of ways in which 10 beads of different colors be arranged on necklace so that particular 4 beads are always together.
- 8640
- 4320
- 720
- 2436
Show the answer
Answer: 8640
Let us consider the 4 beads as one set. Then, we have to arrange this one set and the remaining 6 beads which is equal to 7 in (7-1)! = 720 ways
The 4 beads of the set can be arranged in 4! = 24 ways.
Since in a necklace clockwise or anticlockwise does not make any difference. The total number of permutations = $\frac{720 \text{x} 24}{2}$ = 8640
