Question

How many arrangements can be formed out of the letters of the word EXAMINATION so that vowels always occupy odd places?

(a) 72000

(b) 86400

(c) 10800

(d) 64000

Answer 1

Correct option is (c) 10800

There are 11 letters in the word EXAMINATION of which there are 2A’s, 2I’s.

Also there are 5 consonants and 6 vowels.

The 6 vowels can be arranged in 6 odd places in \(\frac{6!}{2!2!}\) ways (2A’s, 2I’s)

The 5 consonants can be arranged in 5 even possible = \(\frac {5!}{2!} \) ways (2N’s)

\(\therefore\) Total number of arrangements possible = \(\frac{6!}{2! \times 2!} \times \frac {5!}{2!}\)

\(= \frac {6 \times 5 \times 4 \times 3}2 \times 5 \times 4 \times 3\)

\(= 10800\)