Arranging 3 girls and 9 boys 
Three girls A, B and C, and nine boys are lined up in a row. In how many ways this can be done if B must lie between A and C, and A, B must be separated by exactly 4 boys.

I have used the following approach. First select 4 boys ${9 \choose 4}$ and then combine them in between A and B. After that, partition the remaining 5 boys into 3 (similar to r balls into n boxes, here n is $(3-1)$) with ${ 5 + 2 -1\choose 2 - 1} $. Hence, the answer should be: ${9 \choose 4}{ 5 + 2 -1\choose 2 - 1} 4! 5! $. Can anybody tell me if this is correct or not? 
 A: As pointed out in the comments by Barry Cipra, you made two errors.  They are:


*

*$A$ can be to the left or right of $B$, so you have to double your answer.

*The number of ways to distribute $5$ identical objects to $3$ compartments is $\binom{5 + 3 - 1}{3 - 1}$.


With these corrections, your answer becomes
$$2\binom{9}{4}\binom{5 + 3 - 1}{3 - 1}4!5!$$
which is correct.
Method 1:  We modify your approach.  
Choose whether $A$ or $C$ is to the left of $B$.  Suppose it is $A$.  Let $x_1$ be the number of positions to the left of $A$ to be filled with boys; let $x_2$ be the number of positions to the right of $B$ and left of $C$ to be filled with boys; let $x_3$ be the number of positions to the right of $C$ to be filled with boys.  Since $4$ boys must be between $A$ and $B$, we obtain
\begin{align*}
x_1 + 4 + x_2 + x_3 & = 9\\
x_1 + x_2 + x_3 & = 5
\end{align*}
which is an equation in the nonnegative integers.  A particular solution of this equation corresponds to the placement of $3 - 1 = 2$ addition signs in a row of $5$ ones.  For instance,
$$1 1 + 1 1 1 +$$
corresponds to the solution $x_1 = 2$, $x_3 = 3$, $x_3 = 0$.  The number of such solutions is 
$$\binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}$$
since we must choose which two of the seven positions required for five ones and two addition signs will be filled with addition signs.  Once this choice is made, there are $9!$ ways to arrange the boys in the positions chosen for them.  Hence, the number of admissible arrangements is 
$$\binom{2}{1}\binom{7}{2}9!$$
Method 2:  We have seven objects to arrange, a block of length $6$ consisting of girls $A$ and $B$ and four positions for boys who will be placed between them, girl $C$, and five positions for the other boys.  We choose a position for the block in seven ways and a position for girl $C$ in six ways.  If girl $C$ is to the left of the block, girl $A$ must occupy the right end of the block and girl $B$ must occupy its left end.  If girl $C$ is to the right of the block, girl $A$ must occupy the left end of  the block and girl $B$ must occupy its right end.  The boys can be arranged in the nine open positions in $9!$ ways.  Hence, the number of admissible arrangements is 
$$7 \cdot 6 \cdot 9!$$
