Question

Consider the letters of the word $MATHEMATICS$.
The possible number of words in which no two vowels are together is

• A. $7!\cdot {}^8{C}_4\cdot \frac{4!}{2!}$
• B. $\frac{7!}{2!}\cdot {}^8{C}_4\cdot \frac{4!}{2!}$
• C. $\frac{7!}{2!2!}\cdot {}^8{C}_4\cdot \frac{4!}{2!}$
• D. $\frac{7!}{2!2!2!}\cdot {}^8{C}_4\cdot \frac{4!}{2!}$

Answer 1

The correct option is C $\frac{7!}{2!2!}\cdot {}^8{C}_4\cdot \frac{4!}{2!}$

Consonants can be placed in $\frac{7!}{2!2!}$ ways.
Then there are $8$ places and $4$ vowels.
$\therefore$ The number of words
$=\frac{7!}{2!2!}\cdot {}^8{C}_4\cdot \frac{4!}{2!}$