Question

. If the permutation of the word WHITE is selected at random, how
many of the permutations
i. Begins with a consonant?
ii. Ends with a vowel?
iii. Has a consonant and vowels alternating?

Answer 1

(i) begins with a consonant

Ans: 72

First letter can be W or H or T  (3 ways)

Remaining 4 letters can be arranged in 4! ways

Total : 3 × 4! =  72

(ii) ends with vowels

Ans: 48

Last letter can be I or E (2 ways)

Remaining 4 letters can be arranged in 4! ways

Total : 2 × 4! =  48

(iii) has consonants and vowels alternating

Ans: 12

There are 2 vowels and 3 consonants

Arrange the 2 vowels (2! ways)

Now there are 3 spaces to keep the 3 consonants( 3! ways)

Total : 2!×3! = 12