Question

All the letters of the word 'EAMCOT' are arranged in different possible ways. find the number of arrangement in which no two vowels are adjacent to each other.

Answer 1

There are 6 letters in the word 'EAMCOT'. Out of these letters. 'E', 'A' and 'O' are the three vowels.
The remaining three consonants can be arranged in $3_{P_3}$ ways. In each of these arrangements 4 places are created, which are denoted by Cross mark.
i.e., XCXCXCX
Since, no two vowels are to be placed adjacent to each other, so we may arrange 3 vowels in 4 places in $4_{P_3}$ ways.
The total number of arrangements
= ${}^3{P}_3{\times }^4{P}_3$
= $3!\times 4!$ [Since, ${}^n{P}_r=\frac{n!}{(n-r)!}$]
=6 $\times$ 24
= 144.