Consider the word ' $P E R M U T A T I O N$ ' Out of all the permutations in how many words letter $T$ will be in the middle?

10!

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\frac{10!}{2!}

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11!

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None of these

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Solution

The correct option is D $10!$
$T - -- -$

Leftover letters $= 10$
Number of blanks $= 10$
Number of ways of averaging $10$ letters in $10$ blanks
$=^{10} P_{10} = 10!$

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