$\begin{array}{1 1}(A)\;32810\\(B)\;33810\\(C)\;34810\\(D)\;35810\end{array} $

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- $nP_r=\large\frac{n!}{(n-r)!}$

In given word their are $4I,4S,2P$ and $1M$

Total number of permutations with no restriction =$\large\frac{11!}{4!4!2!}$

If take 4I as one letter then total letters become =11-4+1

$\Rightarrow 8$

If P is the permutations when 4i's are not together,then

$P=\large\frac{11!}{4!4!2!}-\frac{8!}{4!2!}$

$\;\;\;=\large\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4 \times 3\times 2 \times 1}{4\times 3\times 2\times 1\times 2\times 1\times 4!}-\frac{8\times 7\times 6\times 5(4!)}{2\times 1(4!)}$

$\;\;\;=34650-840$

$\;\;\;=33810$

Hence (B) is the correct answer.

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