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 × … × a_n ways (a₁ way and a₂ way and a₃ + … + a_n way)

All about Permutation:

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 B_r are alike of r^th kind, such that (B₁ balls + B₂ balls + B₃ balls ----- B_r 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)}P_r ---- (k are the number of objects not included)
b) Always included: ^{(n – k)} C_{r–k x r!} ---- (k is the number of objects always included)

All about combinations:

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 needed	Condition 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) ⁿC_n = ⁿC₀ = 1

2) ⁿC_{n – 1} = ⁿC₁ = n

3) ⁿC_r = ⁿ C_{n – r}

4) 0! = 1

5) n! = n (n – 1)!

6) nPr =	n!
	(n – r)!

7) nCr =	nPr!
	r!

Types of Questions

Q 1. Find the value of ⁵⁰P₂

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

Q 5. Find the value of ²⁰C₁₇

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