Seating arrangements of 7 boys and 5 girls in a row. In how many ways can these boys and girls be arranged in a row if between two particular boys A and B there are no boys but exactly 3 girls?
 A: What follows assumes that you are only interested in the gender distribution.  That is, you don't care if Amy and Clare change places but you do care if Amy and Bob do.  If you do care about the identity of each child, you must multiply by $5!\times 5!$ to get all the possible combinations. (Note: as two of the boys are already singled out, we should only permute the remaining $5$).
Your word either contains $AGGGB$ or $BGGGA$.  
Let's count the words that contain $AGGGB$ (of course the other gives the same count).
There are $8$ places to start $AGGGB$.  
Having located that block, we now have $7$ slots which we must populate with $5$ boys and $2$ girls. There are $\binom 75$ ways to do that.  Thus we get $8\times \binom 75$. Doubling to include the case $BGGGA$ we get $$16\times \binom 75=\fbox {336}$$
A: Take $AGGGB/BGGGA$ as a block, plus $5$ boys, $2$ girls, and permute the $8$ objects
$2\times \dfrac{8!}{5!2!} = 336$
The above assumes that identities other than those of $A$ and $B$ don't matter.
If they do, multiply by $5!5!$ 
A: Assume that we care about where each child sits.
1) If we consider the block containing A and B, we have 2 ways to place A and B in their seats 
$\hspace{.18 in}$and then $5\cdot4\cdot3$ ways to seat the 3 girls in between.
2) We still have to arrange the remaining 5 boys, the other 2 girls, and this block in order, and
$\hspace{.18 in}$ this can be done in $8!$ ways.
Therefore there are $(2\cdot5\cdot4\cdot3)(8!)=4,838,400$ arrangements.
