Re: Combinatorics question

*Hint*: Consider the four I's as a block, then we have seven elements:

$\displaystyle \boxed{IIII}\;MSSSSP,\;\ldots,\; MS\;\boxed{IIII}\; SPSS,\ldots$

We have $\displaystyle \frac{7!}{2!4!1!}$ arrangements of those seven elements. Now, move conveniently the four I's.

Re: Combinatorics question

Hey, thanks for your reply FernanoRevilla. That makes sense. (n.b. I left an extra P out of MISSISSIPPI, but have now edited it in). But I still dont see where the $\displaystyle {8 \choose4}$ comes from. If the four had to remain in a block then the solution would be $\displaystyle {8 \choose 1}$ but if any of the positions can be chosen for any of the Is then some of these will contain adjacent Is and others will not eg. MISISISISPP It seems to me this will be included in the $\displaystyle {8 \choose 4}$ factor. No?

Thanks again, MD

Re: Combinatorics question

Quote:

Originally Posted by

**Mathsdog** IN how many ways can the letters in MISSISSIPPI be arranged so that __no__ two Is are adjacent?

The solution is allegedly $\displaystyle {{8}\choose{4}} { 7! \over 2!4!1!}$

Reply #2, answered a different questiona: "all I's are adjacent'.

For two Is are adjacent, consider "MSSSSPP" that string can be arranged is $\displaystyle \frac{7!}{2!\cdot 4!}$ ways.

Now look at "_M_S_S_S_S_P_P_" there are eight spaces into which the I's can be placed.

Choose four if them: $\displaystyle \binom{8}{4}.$

Re: Combinatorics question

Quote:

Originally Posted by

**Plato** Reply #2, answered a different questiona: "all I's are adjacent'.

No, considering all I's adjacent was a strategy. For that reason I said: now, move conveniently the four I's.

Re: Combinatorics question

Hello, Mathsdog!

Quote:

In how many ways can the letters in MISSISSIPPI be arranged

so that no two I's are adjacent?

The solution is allegedly: $\displaystyle {8\choose4} { 7! \over 2!4!1!}$

Arrange the letters $\displaystyle \{P,P,S,S,S,S,M\}$ in a row.

. . There are: $\displaystyle {7\choose2,4,1} \:=\:\frac{7!}{2!\,4!\,1!}$ arrangements.

Insert a space before, after and between the seven letters:

. . $\displaystyle \square\;X\;\square\;X\;\square\;X\;\square\;X\; \square \;X\;\square\;X\;\square\;X\;\square$

Choose 4 of the 8 spaces and insert the I's.

. . There are:.$\displaystyle {8\choose4}$ choices.

Therefore, there are:.$\displaystyle {8\choose4}\,\frac{7!}{2!\,4!\,1!}$ ways.