Counting arrangements of Mississippi in which the first S precedes the first I, and the first I precedes the first P How many ways are there to arrange the letters in MISSISSIPPI such that the first S precedes the first I, and the first I precedes the first P? 
I get that to arrange the first S before I there are 5!/3!2! ways, but when I go to distribute the ways to do the I before P, I would have already counted the I's.
I am literally at a loss for this problem? 
 A: Consider the first S, I and P
_ S _ I _ P _
The second P can only go in the last gap in $\binom{1+0}{0}$ ways.
_ S _ I _ P _ P _
The 3 remaining Is can only go in the last 3 spaces in $\binom{3+2}{2}$ ways. e.g.
_ S _ I _ I _ P _ I _ P _ I _
The 3 remaining Ss can only go in the last 7 spaces for each in $\binom{3+6}{6}$ ways. e.g.
_ S _ S _ I _ S _ I _ P _ I _ P _ S _ I _
The single M can go in any of the 11 spaces for each in $\binom{1+10}{10}$ ways, hence
$$\text{desired count}=\binom{1}{0}\binom{5}{2}\binom{9}{6}\binom{11}{10}=9240\qquad\blacksquare$$
A: We will arrange the S's and I's first, position the two P's, then place the M last.
We place an S in the first position.  $$S\square\square\square\square\square\square\square$$
Since three of the remaining seven positions must be filled by an S, there are $$\binom{7}{3}$$ arrangements of four S's and four I's in which the first S appears before the first I.  
Since we need to consider where the first P is relative to the first I, we need to consider cases.  
Case 1:  The first I is in the second position. 
$$SI\square\square\square\square\square\square$$
There are $$\binom{6}{3}$$ such arrangements since three of the final six positions must be filled with an S.  We must place the two P's in one or of the seven positions indicated by a wedge.
$$SI\wedge\square\wedge\square\wedge\square\wedge\square\wedge\square\wedge\square\wedge$$  The number of ways we can do this is the number of solutions of the equation 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 2$$
where $x_k$ represents the number of P's in the $k$th position.  The number of solutions of this equation is equal to the number of ways six addition signs can be placed in a row of two ones.  For instance, 
$$+ + 1 + + + 1 +$$
corresponds to the solution $x_1 = x_2 = 0$, $x_3 = 1$, $x_4 = x_5 = 0$, $x_6 = 1$, and $x_7 = 0$, while
$$+ + + + + + 1 1$$
corresponds to the solution $x_1 = x_2 = x_3 = x_4 = x_5 = x_6 = 0$ and $x_7 = 2$.  The number of such solutions is 
$$\binom{2 + 6}{6} = \binom{8}{6} = \binom{8}{2}$$
since we must choose which six of the eight symbols (six addition signs and two ones) will be addition signs.  Hence, in this case, there are 
$$\binom{6}{3}\binom{8}{2}$$
arrangements of four S's, four I's, and two P's in which the first S occurs before the first I and the first I appears before the first P.  
Case 2:  The first I is in the third position.
$$SSI\square\square\square\square\square$$ 
There are 
$$\binom{5}{2}$$
such arrangements since two of the final five positions must be filled with an S.  We must place the two P's in one or more of the six positions indicated by a wedge.
$$SSI\wedge\square\wedge\square\wedge\square\wedge\square\wedge\square\wedge$$
The number of ways this can be done is equal to the number of solutions of the equation 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 2$$
which is

 $$\binom{5 + 2}{5} = \binom{7}{5} = \binom{7}{2}$$

Hence, in this case, the number of arrangements of four S's, four I's, and two P's in which the first S appears before the first I and the first I appears before the first P is 

 $$\binom{5}{2}\binom{7}{2}$$

Case 3:  The first I is in the fourth position.
$$SSSI\square\square\square\square$$
There are 
$$\binom{4}{1}$$
such arrangements since we must place an S in one of the last four positions. The two P's must be placed in one of the five positions indicated by a wedge.
$$SSSI\wedge\square\wedge\square\wedge\square\wedge\square\wedge$$
The number of ways this can be done is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 2$$
in the nonnegative integers, which is 

 $$\binom{2 + 4}{4} = \binom{6}{4} = \binom{6}{2}$$

In this case, the number of arrangements of four S's, four I's, and two P's in which the first S appears before the first I and the first I appears before the first P is 

 $$\binom{4}{1}\binom{6}{2}$$

Case 4:  The first I is in the fifth position.
$$SSSSIIII$$
There is only one such arrangement.  The two P's must be placed in one or more of the four positions indicated by a wedge.
$$SSSSI\wedge I\wedge I \wedge I \wedge$$
The number of ways we can do this is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 2$$
in the nonnegative integers, which is 

 $$\binom{3 + 2}{3} = \binom{5}{3} = \binom{5}{2}$$

Total: Since these cases are exhaustive and mutually exclusive, the number of ways of arranging four S's, four I's, and two P's such that the first S appears before the first I and the first I appears before the first P is

 $$\binom{6}{3}\binom{8}{2} + \binom{5}{2}\binom{7}{2} + \binom{4}{1}\binom{6}{2} + \binom{3}{0}\binom{5}{2}$$

For each permissible arrangement of the four S's, four I's, and two P's, we can place the M in one of eleven positions, namely the nine positions between successive letters and the two ends of the row.  Hence, there are 

 $$\binom{11}{1}\left[\binom{6}{3}\binom{8}{2} + \binom{5}{2}\binom{7}{2} + \binom{4}{1}\binom{6}{2} + \binom{3}{0}\binom{5}{2}\right]$$

arrangements of the letters of MISSISSIPPI in which the first S appears before the first I and the first I appears before the first P. 
