Letter permutation MISSISISPPI, S come before any I

Question

If there is no restriction, the number of ways to organize letter of MISSISSIPPI is, $$ \frac{11!}{4!4!2!} $$

The restriction is,
all Ss come before any Is. So I group both letters, then there are total 4 letters. (PP,M, (Ss and Is)). $$ \frac{4!}{2!}$$.

Is it the correct way?

Answer 1

No, it is not. The problem merely states that every $S$ appears before any $I$, not that the $S$s have to appear together or that the $I$s have to appear together.

There are eleven letters in $MISSISSIPPI$, so we have eleven positions to fill. We can place the $M$ in eleven ways, which leaves ten open positions. We choose two of them for the $P$s. Once we have done that, there is only one way to fill the remaining positions with the $S$s and $I$s since every $S$ must appear before the first $I$. Hence, there are $$\binom{11}{1}\binom{10}{2}$$ admissible arrangements.

Answer 2

Since all S should come before I consider the word SSSSIIII. Now we need to add the letters MPP. First letter we add can take 9 positikns (anywhere between letters or at ends of SSSSIIII). Second and third can be added at 10 and 11 positions. But P appears twice so we divide by 2!. Final answer is (11!/8!)/2!