Question

Consider the letters of the word $MATHEMATICS$.
Possible number of words taking all letters at a time such that in each word both $M$’s are together and both $T$’s are together but both $A$’s are not together is

• A. $\frac{11!}{2!2!2!}-\frac{10!}{2!2!}$
• B. $7!\times {}^8{C}_2$.
• C. $\frac{6!4!}{2!2!}$
• D. $\frac{9!}{2!2!2!}$

Answer 1

The correct option is B $7!\times {}^8{C}_2$.

Make a group of both $M$'s and another group of $T$'s. Then except $A$'s we have $5$ letters remaining. So $M$'s, $T$'s, and the letter except $A$'s can be arranged in $7!$ ways.
Therefore, total arrangements is $7!\times {}^8{C}_2$.