How many different permutations are there of the sequence of letters in "MISSISSIPPI"? There are 11 letters in the word. 
M - 1
I - 4
S - 4
P - 2
so the number of different permutations is $\dfrac{11!}{1! 4 !4!2!}$
Is this correct solution?
 A: Yes! Here's why it's correct:
If you numbered each of the $I$s, $S$s, and $P$s, there would be 11! permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P_1, P_2).$$

But, because you don't care about the numbers of the $P$s, each permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ is actually double-counted, once with $P_1$ before $P_2$ and once with $P_2$ before $P_1$. For example, in the original 11! permutations $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_1,P_2,I_1$$ looks the same as $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_2,P_1,I_1$$ if we ignore the numbers of the $P$s.
So, $\frac{11!}{2!}$ is the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P).$$

For the same reason, we find 24-times the number of permutations of
$$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$
as
$$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P),$$
one for each of the $4! = 24$ orderings of $(I_1, I_2, I_3, I_4).$ Thus there are $\frac{11!}{2!4!}$

And again there are 24-times the number of permutations of
$$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P)$$
as
$$(M, I, I, I, I, S, S, S, S, P, P),$$
one for each of the $4! = 24$ orderings of $(S_1, S_2, S_3, S_4).$ Thus there are $\frac{11!}{2!4!4!}$ total.
A: An alternative solution:


*

*Choose $1$ out of $11$           places for the M

*Choose $4$ out of $10$ remaining places for the I's

*Choose $4$ out of $ 6$ remaining places for the S's

*Choose $2$ out of $ 2$ remaining places for the P's



$$\binom{11}{1}\cdot\binom{10}{4}\cdot\binom{6}{4}\cdot\binom{2}{2}$$
Please note that you can apply this process in any order you'd like.
