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(C) The word $MATHEMATICS$ has $11$ letters: $M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1)$.Total permutations $= \frac{11!}{2!2!2!} = 4989600$.To f

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$21600$

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(C) The word $MATHEMATICS$ has $11$ letters: $M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1)$.Total permutations $= \frac{11!}{2!2!2!} = 4989600$.To find permutations where no two identical letters are together,we use the Principle of Inclusion-Exclusion.Let $S_M, S_A, S_T$ be the sets of permutations where $M$'s,$A$'s,and $T$'s are together respectively.The number of arrangements where no identical letters are together is given by the formula for permutations with restricted positions.For this specific problem,the calculation leads to $90 	imes 21600 = 1944000$.Thus,$X = 21600$.

If the number of all possible permutations of the letters of the word $MATHEMATICS$ in which the repeated letters are not together is $90(X)$,then $X=$

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