How To Solve Quickly Permutation Combination

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 n^r = 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 n^r = 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