MISSISSIPPI with 2 constraints This is a post with two questions:
The problem states:
How many arrangements of MISSISSIPPI are there in which the first I precedes the first S and the first S precedes the first P.
First answer and question:
For the $S$'s the basic construction is:
$$
I \wedge S \wedge P \wedge P \wedge,
$$
so there are 4 spaces where to put 3 $S$'s: $\binom{6}{3}$.
For each arrangement of the $S$'s there are $8$ spaces where to fit $3$ $I$'s: $\binom{10}{3}$.
Finally, for the $M$, there are 11 ways to put it. Answer: $\binom{6}{3}\binom{10}{3}\binom{11}{1}=26400$. Apparently this is wrong, is it?

Second answer and question
The basic construction is 
$$
\wedge I \wedge S \wedge P \wedge P \wedge,
$$
but here, the S's can occupy places $2-5$, $I$'s and the M can occupy all the places.
(a) There's zero $I$'s in places $2-5$ and $M$ is also in place $1$:
$1 \binom{6}{3}\frac{4!}{3!}$. (The first one is arrange the $S$'s, the arrange them in the allowed places, and the last term is to arrange $I$'s and the $M$ in the first place.
(b) There's zero $I$'s in places $2-5$ and $M$ is also in places $2-5$:
$\frac{4!}{3!} \binom{7}{4} 1$. (Again, the first term is to arrange the $S$'s and the $M$, then place them in the places $2-5$, the last term is the arrangement of $3$ identical I's.
(c) $1$ $I$ in places $2-5$ and $M$ in place $1$: $\frac{4!}{3!}\binom{7}{4}\frac{3!}{2!}$
(d) $1$ $I$ in places $2-5$ and $M$ in places $2-5$: $\frac{5!}{3!}\binom{8}{5} 1$
(e) $2$ $I$'s in places $2-5$ and $M$ in place $1$: $\frac{5!}{3!2!}\binom{8}{5} 2!$
(f) $2$ $I$'s in places $2-5$ and $M$ in places $2-5$: $\frac{6!}{3!2!}\binom{9}{6}$
(g) $3$ $I$'s in places $2-5$ and $M$ in place $1$: $\frac{6!}{3!3!}\binom{9}{6}$
(h) $3$ $I$'s in places $2-5$ and $M$ in places $2-5$: $\frac{7!}{3!3!}\binom{10}{7}$
Summing up all the cases I obtain the same number as my answer 1: $\binom{6}{3}\binom{10}{3}\binom{11}{1}$.
Now, my second question is the following, is it wrong this last argument?
I don't know if I can save my answer because for example I think I have double counting in the following example (I placed in parenthesis the position of the letters that occupy an allowed space):
$$
(M) I (SISI) S \wedge P (IS) P \wedge = \\
(M) I \wedge S (ISIS) P (IS) P \wedge
$$
This is some special case of (g), and they count differently because of the position of the letters and the permutation in the arrangement, but both are
at the end the word $MISISISPISP$.
 A: Answer for first question.  Your method is basically correct.  However, after you have
$$I\cdots S\cdots P\cdots P\cdots\ ,$$
there are three possible places for the remaining $S$s, not four, because the one you have already is the first $S$.  Similarly, when you have
$$I\cdots S\cdots S\cdots P\cdots P\cdots S\cdots S\cdots$$
or something similar, there are seven places, not eight, for the remaining $I$s.  So the number of arrangements is
$$\binom53\binom9311=9240\ .$$
Note that to count the arrangements correctly you must specify at the beginning which is the first $S$ (and $I$ and $P$).  Once you have done this, then obviously you cannot place another $S$ before the first $S$.  If you don't do it this way, then taking for example
$$I\cdots_1S\cdots_2$$
and putting the next $S$ in place $1$ or place $2$ will give you the same arrangement counted twice.  
A: Here is an alternate solution:
Since there are $11$ letters in MISSISSIPPI, there are $11$ ways to place the M.  We must fill the remaining $10$ positions with four I's, four S's, and two P's.  Since the first I must appear before the first S and the first S must appear before the first P, we must fill the leftmost open position with an I.  We are left with nine spaces to fill.  We can fill three of them with an I in $\binom{9}{3}$ ways.  This leaves us with six spaces to fill with the four S's and two P's.  Since the first S must appear before the first P, we must fill the leftmost open position with an S.  This leaves us with five spaces to fill.  We can fill three of them with the remaining S's in $\binom{5}{3}$ ways.  We can place the two P's in the two remaining positions in one way.  Hence, the number of permutations of MISSISSIPPI in which the first I appears before the first S and the first S appears before the first P is $$11\binom{9}{3}\binom{5}{3}$$
