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In how many ways can the letters of the word PERMUTATIONS be arranged if the Words start with P and end with $S$?

$\begin{array}{1 1}(A)\;1814400\\(B)\;1804400\\(C)\;1824400\\(D)\;1834400\end{array} $

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Letters between P and S are ERMUTATION .These 10 letters having T two times
These letters can be arranged in $\large\frac{10!}{2!}$ ways
$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 1}$
$\Rightarrow 1814400$ ways
Hence (A) is the correct answer.
answered May 13, 2014 by sreemathi.v

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