# How To Solve Quickly Permutation Combination

## How to Solve Permutation and Combination Questions Quickly

How to solve Permutation and Combination Questions Quickly has been discussed on this page along with basic concepts.

Permutation is basically called as a arrangement where order does matters.Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time.It is represented as,$\mathbf{^nP_r = \frac{n!}{(n-r)! }}$

Combination is basically called as a selection where order does not matters.Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time. It is represented as$\mathbf{^nC_r = \frac{n!}{(n-r)! r! }}$

## Basic concept of Permutation and Combination

### How to Solve Permutation and Combination Questions Quickly.

• Permutation is an arrangement of objects in a definite order.
• Number of all permutations of n things, taken r at a time, is given by
nPr = $\mathbf{\frac{n!}{(n-r)!}}$
• Combination is selection of objects where order does not matter.
• Number of all combinations of n things, taken r at a time, is given by
nCr = $\frac{n!}{(n-r)! r! }$
• Here we can easily understand how to solve permutation and combination easy.

### Type 1: How to Solve Quickly Permutation and Combination Different ways to arrange (with repetition)

Question 1.How many 3 letter words with or without meaning can be formed out of the letters of the word MONDAY when repetition of words is allowed?

Options:

A. 125

B. 216

C. 120

D. 320

Solution:      6 * 6 * 6 = 216

OR

We can solve directly by formula nr = 63 = 216

Correct option: B

Question 2. In how many ways the letters in the word TOOTH can be arranged?

Options:

A. 120

B. 40

C. 20

D. 30

Solution:   $\frac{5!}{2! × 2! }$

= $\frac{5 × 4 × 3 × 2 × 1 }{2 × 1× 2 × 1 }$

= $\frac{120}{ 4}$

= 30

Correct option: D

Question 3.How many three digit numbers can be formed using digits 2, 3, 4, 7, 9 so that the digits can be repeated.

Options:

A. 125

B. 360

C. 24

D. 6

Solution:    Each place can be filled by any one of 5 digits

Total numbers = 5 * 5 *5 =125

OR

We can solve directly by formula nr = 53 = 125

Correct option: A

### Type 2: Different ways to arrange (without repetition)

Question 1. How many five letter words with or without meaning, can be formed from the word ‘COMPLEXIFY’, if repetition of letters is not allowed?

Options:

A. 43200

B. 30240

C. 12032

D. 36000

Solution:    10P5 = $\frac{10!}{(10 – 5)!}$  = 10 * 9 * 8 * 7 * 6 = 30240

Correct option: B

Question 2.In how many different ways can the letters of the word ‘LOGARITHMS’ be arranged so that the vowels always come together?

Options:

A. 6720

B. 241920

C. 40320

D. 360344

Solution:     In such questions we treat vowels as one letter.

So the word becomes LGRTHMS (OAI)

It means there are total 8 letters. Therefore, number of ways of arranging these letters = 8! = 40320

Now, there are three vowels (OAI), number of ways of these letters can be arranged = 3! = 6

Required number of words = 40320 * 6 = 241920

Correct option: B

Question 3.How many three digit numbers can be formed from the digits 3, 4, 5, 7, 8, and 9. Also, the number formed should be divisible by 5 and no repetition is allowed?

Options:

A. 20

B. 24

C. 25

D. 10

Solution:     The number which is divisible by 5 has 5 or 0 at one’s place. In this case we must have 5 at the unit place as 0 is not in the list.

There are total 6 digit out of which last digit is fixed by 5. Therefore, we are left with 5 digits (3, 4, 7, 8, 9) at the tens place.

Similarly, the hundred place can be filled by 4 digits.

So, required number = 4 * 5 * 1 = 20

Correct option: A

### Type 3: How To Solve Permutation and Combination Question- (with repetition)

Question 1.An ice cream seller sells 5 different ice-creams. John wants to buy 15 ice creams for his friends. In how many ways can he buy the ice-cream?

Options:

A. 1450

B. 3768

C. 3876

D. 1540

Solution:     r + n -1Cr = 15 + 5 – 1C15 =19C15

We know that, nCr = $\frac{n!}{(n-r)! r! }$

19C15 = $\frac{19!}{(19-15)! 15! }$= 3876

Correct option: C

Question 2.There are 5 types of soda flavor available in a shop. In how many ways can 10 soda flavors be selected?

Options:

A. 1454

B. 1001

C. 1211

D. 1540

Solution:       r + n – 1Cr = 10 + 5 – 1C10 = 14C10

We know that, $\frac{n!}{(n-r)! r! }$

14C10 = $\frac{14!}{(14-10)! 10! }$ = 1001

Correct option: B

Question 3. In how many ways can 16 identical toys be divide in 4 children?

Options:

A. 966

B. 696

C. 969

D. 996

Solution:      r + n – 1Cr = 16 + 4 – 1C16 = 19C16

We know that, nCr = $\frac{n!}{(n-r)! r! }$

19C16 = $\frac{19!}{(19-16)! 16! }$ = 969

Correct option: C

### Type 4: Permutation and Combination Solve Question Quickly. (without repetition)

Question 1. A wooden box contains 2 grey balls, 3 pink balls and 4 green balls. Fins out in how many ways 3 balls can be drawn from the wooden box. Make sure that at least one pink ball is included in the draw?

Options:

A. 64

B. 46

C. 56

D. 65

Solution:     According to the question, we have, (one pink and two non-pink balls) or (two pink and one non-pink balls) or (3 pink)

Therefore, required number of ways are (3C1 * 6C2) + (3C2 * 6C1) + (3C3) = 45 +18 + 1 = 64

Correct option: A

Question 2.There are 5 boys and 10 girls in a classroom. In how many ways teacher can select 2 boys and 3 girls to make a dance group?

Options:

A. 720

B. 1200

C. 240

D. 840

Solution:     Required numbers of ways = 5C2 * 10C3 = 10 * 120 = 1200

Correct option: B

Question 3.There are 10 consonants and 5 vowels. Out of which how many words of 5 consonants and 2 vowels can be made?

Options:

A. 1270080

B. 120052

C. 210789

D. 720432

Solution:     Number of ways of selecting (5 consonants out of 10) and (2 vowels out of 4) = 10C5 * 5C2 = 252
Number of ways of arranging 7 letters among themselves = 7!

=7 x 6 x 5 x 4 x3 x2 x 1

=5040

Required number of ways = (252 x 5040) = 12,70,080

Correct option: A

Read Also –  Formulas to solve permutation questions