# Permutations and Combinations - Aptitude test, questions, shortcuts, solved example videos

Video on Permutations and Combinations - shortcuts, tips and tricks

## Permutation and Combination

We have always faced the problem of understanding the difference between permutation and combination.

This chapter will definitely clear the concepts of permutation and combination, the only thing you have to do is thoroughly understand the difference between the two terms and as well learn the quick tips to solve problems based on this chapter

Difference between permutation and combination

What is permutation?

Permutation: The various ways of arranging a given number of things by taking some or all at a time are all called as permutations.
Permutation includes word formation, number formation, circular permutation, etc. In permutation, objects are to be arranged in particular order. It is denoted by n P r or P(n, r).

Example: Arrange the given 3 numbers 1, 2, 3 by taking two at a time.
Now these numbers can be arranged in 6 different ways: (12, 21, 13, 31, 23, 32).

Here,

12 and 21, 13 and 31 or 23 and 32 do not mean the same, because here order of numbers is important.

What is combination?

Combination: Each of different groups or selections formed by taking some or all number of objects is called a combination.

Combination is used in different cases which include team/group/committee.

In combination, objects are selected randomly and here order of objects doesn’t matter. It is denoted by n C r or C(n, r)

Example: If we have to select two girls out of 3 girls X, Y, Z, then find the number of combinations possible.

Now only two girls are to be selected and arranged. Hence, this is possible in 3 different ways: (XY, YZ, XZ,).

Here,
You cannot make a combination as XY and YX, because these combinations mean the same.

Quick Tips and Tricks

Factorial n!: It is the product of all positive integers less than or equal to n.
Example: 4! = 4 × 3 × 2 × 1 = 24

Theorem of Counting:

1) Rule of Addition: If a first task is performed in x ways and second task is performed in y ways, then either of the two operations can be performed in (x + y) ways

2) Rule of Multiplication: If a first task is performed in x ways and second task is performed in y ways, then both of the two operations can be performed in (x × y) ways

Suppose n different cakes are done in a 1 , a 2 , a 3 , … a n different ways respectively, independent of each other then:

1) Any one of them can be done in a 1 + a 2 + a 3 + … + a n ways. (a 1 way or a 2 way or a 3 + … + a n way)
2) All of them can be done in a1 × a2 × a3 × … × an ways (a1 way and a2 way and a3 + … + an way)

1) Condition 1: Number of permutations of n things, taken r at a time is given as follows:
 nPr = n (n – 1) (n – 2) (n – 3)……. (n – r + 1) = n! (n – r)!

2) Condition 2: If there are N balls and out of these B1 balls are alike, B2 balls are alike , B3 balls are alike and so on and Br are alike of rth kind, such that (B1 balls + B2 balls + B3 balls ----- Br balls) = N balls.

In such condition,
 Number of permutation of these N balls = N! (B1)! × (B2)! × (B3)! × - - - - - (Br)!

3) Condition 3: If number of permutations of n objects are all taken at a time,
 then, nPr = n! = n! 0!

Important points to remember:

1) If N different objects are to be arranged, then they can be arranged in N! ways.

2) N number of objects can be arranged around a circle in (N – 1)! ways.

3) Sometimes we have to solve problems on permutation considering the condition of Repetition

Repetition: This condition is not used unless specified. (Remember)
Number of permutation of N objects taken r at a time when each selected object can be repeated any number of times is given as:

Number of permutations = n r

4) Restricted Permutation: The number of permutations of n objects taken r at a time in which if k particular objects are:

a) Never included: (n – k)Pr ---- (k are the number of objects not included)
b) Always included: (n – k) Cr–k x r! ---- (k is the number of objects always included)

1) Number of combinations of n objects, taken r at a time is given as follows:
 nCr = n! = n(n – 1) (n – 2)…….to r factors (r!)(n – r)! r!

This example will surely clear the concept!

Hint: In the example discussed below, the confusion related to addition and multiplication of terms will also be cleared.

Example: Suppose there are 12 boys and 8 girls, and we have to select 5 volunteers for a particular task. So we have to find the number of possible selections we can make.

Total students are (12 + 8) = 20 and we have to select 5 volunteers.
 The total number of selections can be made nCr ways = 20C5 n! (r!)(n – r)!

The question may be asked in different ways. 2 different conditions are specified below:

1) Out of 5 volunteers, 3 boys and 2 girls must be present.
2) Boys should be in majority.

Condition 1: 3 boys 2 girls are neededCondition 2: If boys are in majority
Out of 12 boys 3 are selected and out of 8 girls 2 are selected.

(Boys) 12C3 and (Girls) 8C2

Both girls and boys are needed, hence they are multiplied.

12C3 × 8C2
1) If only boys are selected as volunteers: 12C5
2) 4 boys and 3 girls: 12C4 × 8C1
3) 3 boys and 2 girls: 12C3 × 8C2

These will be the 3 possibilities, where boys are in majority.

(12C5 ) or (12C4 × 8C1) or (12C3 × 8C2)

(12C5) + (12C4 × 8C1) + (12C3 × 8C2)

Important Formulae:

1) nCn = nC0 = 1

2) nCn – 1 = nC1 = n

3) nCr = n Cn – r

4) 0! = 1

5) n! = n (n – 1)!
 6) nPr = n! (n – r)!

 7) nCr = nPr! r!

Types of Questions

Type 1: Permutation

Q 1. Find the value of 50P2

a. 4500
b. 3260
c. 2450
d. 1470
View solution

Correct Option: (c)

Hint:

 nPr = n! (n – r)!
Here n = 50 and r = 2
 50! = 50 × 49 × 48 = 50 × 49 = 2450 (50 – 2)! 48!

Value of 50P2 = 2450

Q 2. How many words can be formed by using letters of the word ‘DELHI’?

a. 50
b. 72
c. 85
d. 120
View solution

Correct Option: (d)

Hint:
The word ‘DELHI’ contains 5 letters
Therefore, required number of words = 5P5 = 5! = (5 × 4 × 3 × 2 × 1) = 120
120 words can be formed by using letters of the word ‘DELHI’

Q 3. Find the number of ways the letters of the word ‘RUBBER can be arranged?

a. 450
b. 362
c. 250
d. 180
View solution

Correct Option: (d)

Hint:
The word ‘RUBBER’ contains 6 letters: 2R, 2B, 1 U, 1 E
Therefore,

 The required Number of ways: N! (2R!) × (2B!) × (1U!) × (1E!)
 = 6! (2 × 1) × (2 × 1) × (1) × × (1)
 = 6 × 5 × 4 × 3 × 2 × 1 4
= 6 × 5 × 3 × 2
= 180

Q 4. Find in how many different ways, the letters of the word ‘LEADING’ can be arranged in such way that the vowels always come together?

a. 548
b. 426
c. 720
d. 790
View solution

Correct Option: (c)

The letters in the word ‘LEADING’ are 7.
Number of vowels = EAI
Number of consonants: LDNG
Condition: Vowels always come together. Therefore, let’s consider the vowels as one group ‘EAI’
Hence, vowels ‘EAI’ and consonants LDNG together form 5 letters.
Therefore, these 5 letters can be arranged in 5! Ways = 5! = 5 × 4 × 3 × 2 = 120 ways
Letters in group of vowels ‘EAI’ can also be interchanged. So, these vowels can be arranged in 3! Ways = 6 ways.
All together, required number of ways = 120 x 6 = 720 ways

Type 2: Combination

Q 5. Find the value of 20C17

a. 1260
b. 1140
c. 2580
d. 3200
View solution

Correct Option: (b)

Hint:

 nCr = nPr! r!
 nCr = n! (r!) (n – r)!
 20C17 = 20! (17!) (20 – 17)!
 20C17 = 20 × 19 × 18 × 17! (17!) (3)!
 20C17 = 20 × 19 × 18 3 × 2 × 1
20C17 = 1140

Q 6. Out of 5 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

a. 60
b. 1200
c. 5230
d. 7200
View solution

Correct Option: (d)

Hint:

 nCr = n! (r!) (n – r)!

3 consonants out of 5 can be selected in 5C3 ways
2 vowels out of 4 can be selected in 4C2 ways
In 5C3 and 4C2 ways the consonants and vowels can be selected.
Remember: 5C3 and 4C2, hence multiply the terms 5C3 x 4C2
 5C3 × 4C2 = 5! × 4! = 5 × 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1 = (5 × 2) × (3 × 2) 3! (5 – 3)! 2! (4 – 2)! 3 × 2 × 2 2 × 2

60 groups can be made each having 3 consonants and 2 vowels
Each word contains 3 consonants and 2 vowels i.e 5 letters.
Hint:
If N different objects are to be arranged, then they can be arranged in N! ways.
5 letters can be arranged in 5! Ways.
5! = 5 x 4 x 3 x 2 x 1 = 120
Required number of words = (120 x 60) = 7200

Q 7. A bag contains 2 white marbles, 3 black marbles and 4 red marbles. Find in how many ways, 3 marbles can be drawn, so that at least one black marble is included in each draw?

a. 64
b. 52
c. 58
d. 36
View solution

Correct Option: (a)

Hint:

 1) nCr = n! (r!) (n – r)!

2) nCn = nC0 = 1

Among 3 marbles, at least one black marble is to be drawn. Therefore, the number possible combinations are:
1) 1 black and 2 other colors [3C1 × 6C2] ----- (3C1 and 6C2 because 3 balls are drawn)
2) 2 black and 1 of other color [3C2 × 6C1]
3) All 3 are black marbles [3C3]

Hence, the required number of ways = [3C1 × 6C2] + [3C2 × 6C1] + [3C3] ------ ([3C1 × 6C2] and [3C2 × 6C1] are added because they are number of possibilities “This or that”)
 = 3! × 6! + 3! × 6! + 1 (3 – 1)! (6 – 2)! (2)! (3 – 2)! (2)! (6 – 1)!
 = 3 × 2 × 6 × 5 × 4 × 3 × 2 × 1 + 3 × 2 × 1 × 6 × 5 × 4 × 3 × 2 × 1 + 1 2 (4 × 3 × 2 × 1) (2) 2 5 × 4 × 3 × 2 × 1
= [3 × 15] + [3 × 6] + 1
= 64
3 marbles can be drawn from the bag, with at least one black marble in 64 ways.

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