The 11 number of arrangements that can be made with the letters of the word 'MATHEMATICS' in which all vowels come together, is

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$\frac{8! \times 4!}{2! 2!}$

$\frac{8!}{4! 2! 2!}$

$\frac{8! \times 4!}{2! 2! 2!}$

$\frac{8!}{2! 2! 2!}$

answer is B.

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Detailed Solution

There are 4 vowels viz. A, E, A, I. Considering these four vowels as one letter we have 8 letters (M, T, H, M, T .. C, S and one letter obtained by combining all vowels), out or which M occurs twice, T occurs twice and the rest all different. These 8 letters can be arranged in $\frac{8!}{2! \times 2!}$ ways.

But, the four vowels (A, E, A, I) can be put together in $\frac{4!}{2!}$ ways.

Hence, the total number of arrangements in which vowels are alwavs together $= \frac{8!}{2! \times 2!} \times \frac{4!}{2!}$ .