Question

Consider the letters of the word MATHEMATICS. The possible number of words

• A. when no two vowels are together is $\frac{7!}{2!2!}{}^8{C}_4\frac{4!}{2!}$
• B. when both $M$'s are together and both $T$'s are together but both $A$'s are not together is $28\times 7!$
• C. when all vowels are together is $\frac{8!4!}{2!2!2!}$
• D. when all consonants are together is $\frac{5!7!}{2!2!2!}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A when no two vowels are together is $\frac{7!}{2!2!}{}^8{C}_4\frac{4!}{2!}$
B when both $M$'s are together and both $T$'s are together but both $A$'s are not together is $28\times 7!$
C when all vowels are together is $\frac{8!4!}{2!2!2!}$
D when all consonants are together is $\frac{5!7!}{2!2!2!}$

When no two vowels are together.
Consonants can be placed in $\frac{7!}{2!2!}$ ways. Then there are $8$ places and $4$ vowels. Therefore number of ways is $\frac{7!}{2!2!}{}^8{C}_4\frac{4!}{2!}$

When both M's are together and both T's are together but both A's are not together.
Make one group of both M's and another group of both
T's. Then except A's we have 5 letters remaining. So M's, T's and the letters except A's can be arranged in $7!$ ways. Therefore, total number of arrangements is $7!{\times }^8{C}_2=28\times 7!$

When all vowels are together.
Make one group of all vowels.
The total ways of arrangement is $=\frac{8!\times \frac{4!}{2!}}{2!2!}$

When all consonants are together.
Make one group of all consonants.
The total ways of arrangement is $=\frac{5!\times \frac{7!}{2!2!}}{2!}$