Question

The number of different words that can be formed from the letters of the word $INTERMEDIATE$ such that two vowels never come together is

Answer 1

Given: $INTERMEDIATE$ word contains 12 letters.

Number of vowels = $6(A,E,E,E,I,I)$

Number of consonants $=6(N,T,R,M,D,T)$

First arrange $6$ consonants in which $2(T,T)$ are alike in ways. $2!$

Arrangements of these $6$ consonants creates $(6+1=7)$ gaps.

Solution:

To arrange vowels in these gaps first we need to select $6$ gaps which can be done in = ${}^7{C}_6$ ways.

In 6 vowels, we have $3E^{\prime }$s and $2I^{\prime }s$ & $1A$.

So, $6$ vowels can be arranged in these $6$ selected positions. $\frac{6!}{2!3!}$ ways at

$\therefore$ Total number of required ways = (Arrangement of consonants)$\times$ (Number of ways of selecting 6 places between gaps)
$\times$ (Number of ways of arrangement of 6 vowels at these 6 positions)

= $\frac{6!}{2!}\times {}^7{C}_6\times \frac{6!}{2!3!}$

= $7!\times 30$

= $151200$ ways

Hence, the number of different words that can be formed from the letters of the word $INTERMEDIATE$ such that two vowels never come together is $151200$.