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In how many ways can you arrange all letters in the word MISSISSIPPI so that

1) all four I’s are together?

2) two P’s are NOT next to each other?

3) no two I’s are next to each other?

Brendan
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  • Part 3 is done here, except they use S instead of I. http://math.stackexchange.com/questions/697668/mississippi-combinations-with-the-ss-separated Also here: http://math.stackexchange.com/questions/111828/some-questions-about-combinations/162436#162436 –  Oct 05 '14 at 21:24
  • For 1, the symmetric group on 11 letters acts on the set of arrangements of these 11 letters. The stabilizer of any four letters is a copy of the symmetric group on 7 letters. A consecutive arrangement of four I's is determined by the location of the first I (there are 8 possible choices) and by a permutation on 4 letters. Hence there should be about $(11!/7!)\cdot 8\cdot 4!=1520640$ such arrangments. This is assuming two copies of a single letter are different; if not, scrap the copy of $\Sigma_4$ and some of the stuff in $\Sigma_7$. – Eivind Dahl Oct 05 '14 at 21:59
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    Please show what you have done so far on the problem. – user84413 Oct 05 '14 at 22:51

2 Answers2

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1) Treat IIII as one unit (so you're only arranging 8 objects rather than 11). This would be $8!/(4!2!)$ because of the repeated S's and P's.

2) Use complementary counting. First find the total ways to arrange the letters in MISSISSIPPI. This would be $11!/(4!4!2!)$. Then subtract from that number all the ways the 2 P's are together to get the ways the P's are NOT together. This would be done by again treating PP as one unit, so the total number of arrangements would be $10!/(4!4!)$. The answer would be whatever $11!/(4!4!2!) - 10!/(4!4!)$ turns out to be.

Sheheryar Zaidi
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YLZ
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1) All four I's appear together (as a block) in $\frac{8!}{2! 4!}$ ways. [Division by $4!$ & $2!$ is for $2$ P's & $4$ S's are repeating here.]

2) The two P's are not next to each other in $\frac{11!}{4! 4! 2!}-\frac{10!}{4! 4!}$ ways. [Here $\frac{11!}{4! 4! 2!}$ occurs because $4$ S's $4$ I's & $2$ P's are permuting between them & $\frac{10!}{4! 4!}$ occurs because $4$ S's & $4$ I's are permuting between them.]

hardmath
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Ri-Li
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – hardmath Oct 05 '14 at 22:21
  • Sorry I can't get your point. Will you explain? – Ri-Li Oct 05 '14 at 22:39
  • Currently I view your Answer as being the upper one of two, so the reference to "similar kind of reasons as above" is unclear. In giving an Answer for routine exercises, future Readers will benefit more from a clear exposition of method than from a succinct statement of a final value. – hardmath Oct 05 '14 at 22:47
  • Ok you are right. I am giving it. – Ri-Li Oct 05 '14 at 22:53