Question

Consider the letters of the word 'MATHEMATICS'. Possible number of words taking all letters at a time such that in each word both $M^{\prime }s$ are together and both $T^{\prime }s$ are together but both $A^{\prime }s$ 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 $M^{\prime }s$ and another group of $T^{\prime }s$.

then except $A^{\prime }s$ we have $5$ letters remaining.
So $M^{\prime }s,T^{\prime }s,$ and the letters except $A^{\prime }s$ can be arranged in $7!$ ways.
Therefore, total number of arrangements is $7!{\times }^8{C}_2$.