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There are 11 letters in the word.

M - 1 I - 4 S - 4 P - 2

so the number of different permutations is $\dfrac{11!}{1! 4 !4!2!}$

Is this correct solution?

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    $\begingroup$ Yes, this is exactly it. $\endgroup$ – Jimmy R. Mar 23 '15 at 20:46
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Yes! Here's why it's correct:

If you numbered each of the $I$s, $S$s, and $P$s, there would be 11! permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P_1, P_2).$$


But, because you don't care about the numbers of the $P$s, each permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ is actually double-counted, once with $P_1$ before $P_2$ and once with $P_2$ before $P_1$. For example, in the original 11! permutations $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_1,P_2,I_1$$ looks the same as $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_2,P_1,I_1$$ if we ignore the numbers of the $P$s.

So, $\frac{11!}{2!}$ is the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P).$$


For the same reason, we find 24-times the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ as $$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P),$$ one for each of the $4! = 24$ orderings of $(I_1, I_2, I_3, I_4).$ Thus there are $\frac{11!}{2!4!}$


And again there are 24-times the number of permutations of $$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P)$$ as $$(M, I, I, I, I, S, S, S, S, P, P),$$ one for each of the $4! = 24$ orderings of $(S_1, S_2, S_3, S_4).$ Thus there are $\frac{11!}{2!4!4!}$ total.

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An alternative solution:

  • Choose $1$ out of $11$ places for the M
  • Choose $4$ out of $10$ remaining places for the I's
  • Choose $4$ out of $ 6$ remaining places for the S's
  • Choose $2$ out of $ 2$ remaining places for the P's

$$\binom{11}{1}\cdot\binom{10}{4}\cdot\binom{6}{4}\cdot\binom{2}{2}$$

Please note that you can apply this process in any order you'd like.

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  • $\begingroup$ I think it might be worth explicitly pointing that the result you get is the same as OP's and the ones on other answers. $\endgroup$ – Wojowu Mar 23 '15 at 21:32
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    $\begingroup$ @Wojowu: Isn't that obvious from the opening statement an alternative solution? $\endgroup$ – barak manos Mar 23 '15 at 21:34

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