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This question already has an answer here:

How many anagrams of the word mississippi are there that have at least two consecutive i’s? My approach was: Finding the total amount of anagrams (11! / 4! 4! 2!). And then to calculate the amount of options without 2 consecutive i's. In the end, to find the difference between them. However, the result felt me to big intuitively. Do I miss something?

Thanks

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marked as duplicate by polfosol, Yanior Weg, ThorWittich, Adrian Keister, G-man Sep 17 at 1:05

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  • $\begingroup$ Well, you have a typo. You mean ${11!\over 4!4!2!}$ Is this what's making your answer too big? $\endgroup$ – saulspatz Sep 15 at 18:14
  • $\begingroup$ You are correct. it's 4!4!2!. Editing $\endgroup$ – Bob_Bobb Sep 15 at 18:54
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There is a way to do this, actually.

First, order the letters MSSSSPP. This can be done in $\displaystyle \frac{7!}{4!2!}=105$ ways.

Now, you have $8$ spaces between those letters to possibly insert I's (ends included).

Now, choose $4$ of those $8$ spaces to insert an $I$.

This gives a total of $105\cdot 70=7350$ ways.

Now, we are looking for the complement, so our answer is $\displaystyle \frac{11!}{4!4!2!}-7350=34650-7350=27300$ ways.

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    $\begingroup$ Thanks! Great explanation! $\endgroup$ – Bob_Bobb Sep 15 at 18:57

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