Question

Consider the letters of the word ${}^{\prime }MATHEMATIC{S}^{\prime }$.
The possible number of words that can be formed by 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. $28\times 7!$
• C. $\frac{6!4!}{2!2!}$
• D. $\frac{9!}{2!2!2!}$

Answer 1

The correct option is B $28\times 7!$

Make one group of both $M$'s and another group of both
$T$'s.
Arrange all of them except the $A^{\prime }s$
The number of ways $=7!$
Using gap method, the total number of such words
$=7!\times \times {}^8{C}_2=28\times 7!$