Question

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

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

Answer 1

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

Consonants have 8 gaps when arranged in a row. Consonants can be arranged in $\frac{7!}{2!2!}$ ways. Then there are $8$ places and $4$ vowels
Therefore, number of ways is $\frac{7!}{2!2!}{}^8{C}_4\frac{4!}{2!}$.