Distributions and arrangements

Question

In how many ways can the letters of MISSISSIPPI be arranged so that no two I's are consecutive?

We know we have $4$ I's and each has $11 - 4 + 1 = 8$ 'boxes' to be placed in. This can be done in $\dbinom{8-1}{4-1} = \dbinom{7}{3} = 35$.

Next, we have $7$ remaining letters, which can be arranged in $7!$ ways, and finally, we divide by $4\times 2$ as we have $4$ S's and $2$ P's. This gives us $$\frac{\dbinom{7}{3}\times 7!}{4\times 2} = \frac{35\times 7!}{8} = 22050.$$ (Which is incorrect of course).

Answer 1

I see two problems:

• There are $8$ "boxes" (positions relative to the other numbers) and the only restriction is that you can't put two I's in the same box, so the number of options is $\binom 84$, not $\binom 73$.
• You need to divide by $4!\times2!$, not $4\times 2$.