# How many distinct arrangements of the letters in the word 'Mississippi' have no S's in the first six places?

CONTEXT: Question made up by uni lecturer.

How many distinct arrangements of the letters in the word 'Mississippi' have no S's in the first six places?

My first attempt of the question involved treating the S's as a single block of letters (i.e. you find arrangements of elements of $$\{M, I, I, I, I, P, P, SSSS\}$$), however, this doesn't work because if the first six places can't contain S's, then the last five places will, which means the S's could be separated (i.e. in 'PIMIPISSISS', the block of S's is separated by a 'I').

My second attempt involved subtracting the complement (all distinct arrangements with S's only in the first six places) from the total number of distinct arrangements which is $$\frac{11!}{4!4!2!}=34650$$, but I realised this would require a lot more calculations than just considering the original case.

I'm a bit stuck on how to approach this question; any guidance would be greatly appreciated.

• The last five letters, unordered, have to consist of the four $S's$ plus one other letter, either $M,I$ or $P$ of course. So you have three cases to consider. – lulu May 27 '19 at 10:06

$$\binom{5}{4}\binom{7}{4}\binom{3}{2}\binom{1}{1}$$
• A particular arrangement of the letters of the word MISSISSIPPI is completely determined by which letter is placed in which position. There are $\binom{11}{4}$ ways to choose the positions of the I's, $\binom{7}{4}$ ways to choose four of the remaining positions for the S's, $\binom{3}{2}$ ways to choose two of the remaining three positions for the P's and one way to place the M. That gives $$\binom{11}{4}\binom{7}{4}\binom{3}{2}\binom{1}{1} = \frac{11!}{7!4!} \cdot \frac{7!}{4!3!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!0!} = \frac{11!}{4!4!2!1!}$$ distinguishable arrangements as you found. – N. F. Taussig May 27 '19 at 23:16