Five girls and eleven boys are to be lined up in a row such that... Where is my mistake? 
Five girls and eleven boys are to be lined up in a row such that from
  left to right, the girls are in the order $g_1g_2g_3g_4g_5$. In how
  many ways can this be done if $g_1$ and $g_2$ are to be separated by
  atleast 3 boys, and there is at most one boy between $g_4$ and $g_5$.

First take a look at this which shows the number of ways to put $r$ distinct objects into $n$ distinct boxes such that order matters: If you are too lazy to read it, it basically says that the number of ways to put $r$ distinct objects into $n$ distinct boxes such that the order in the boxes matters is given by $P^{n+r-1}_{r}$ 
The whole idea is to count the arrangements where there is an arbitrary number of boys between $g_1$ and $g_2$ and atmost one between $g_4$ and $g_5$ and then subract the number of arrangements where there are less than $3$ boys between $g_1$ and $g_2$ (atmost one between $g_4$ and $g_5$).
First count the arrangements where there is an arbitrary number of boys between $g_1$ and $g_2$ and atmost one between $g_4$ and $g_5$. If you arrange the girls as $\_g_1\_g_2\_g_3\_g_4g_5\_$, then the dashes can be treated as boxes where to put the boys. Observe that there is no dash between $g_4$ and $g_5$. Case 1: there is no object between $g_4$ and $g_5$. There are $P^{15}_{5}$ such arrangements. Case 2: There is one object between $g_4$ and $g_5$. There are $11\times P^{14}_{5}$ such arrangements. Thus, the total is $3003000$.
Now, count the number of permutations where there are less than $3$ boys between $g_1$ and $g_2$. Case 1: $0$ objects between them. If there is no object between $g_4$ and $g_5$ you have $P^{14}_{4}$ and if there is you have $11\times P^{13}_{4}$, so in this case there are $212784$ arrangements. In the same manner for the cases where there are $1$ and $2$ objects between $g_1$ and $g_2$ you get, respectively, the values $11\times P^{13}_{4} + 11\times 10\times P^{12}_{4} = 1,519,684$ and $2\times {11\choose2}\times P^{12}_4 + 2\times{11\choose2}\times 9\times P^{11}_{4} = 3,920,400$ which alone is larger than $3003000$, so there is obviously a mistake.
I know there are more elegant approaches to this (share them if you like), but the main question is where is the mistake in the current approach?
 A: I don't understand why you claim in your Case 1 that there are $P^{15}_5$ arrangements where the girls are in the given order and $g_4$ is right next to $g_5$. The quantity $P^{15}_5$ is the number of ways to choose an ordered subset of $5$ from a set of $15$, and that doesn't seem to apply here-- you know the five girls will be in the given order.
The way I understand it, you have a line with $15$ spaces, and you want to choose $4$ of those spaces to put your girls (since $g_4$ and $g_5$ come together as a package.) Then the remaining $11$ spaces will be filled as a permutation of the boys. So you have $\binom{15}{4} \cdot 11!$ possible arrangements under those conditions. Then in the other case, pick a boy to go between $g_4$ and $g_5$. Then there are really ony $14$ spaces in the line, so you have $\binom{14}{4}$ choices of spaces to put the girls, and the boys fill the remaining spaces as a permutation again, so Case 2 yields $11 \cdot \binom{14}{4} \cdot 10! = \binom{14}{4} \cdot 11!$.
To solve the problem, first fix three boys to go directly in front of $g_1$ in a given order. There are $\frac{11!}{8!} = 990$ ways to do so. Then use a similar argument as above: In one case, $\binom{12}{4}$ ways to choose places for the girls (with their attachments), and $8!$ ways to fill the remaining places. In the other case, $\binom{11}{4}$ choices of places for girls, times $8!$. The total is $990 \cdot 8! \cdot (\binom{12}{4} + \binom{11}{4})$.
This can be simplified nicely: $990 \cdot 8! = 11 \cdot 10 \cdot 9 \cdot 8! = 11!$, and $\binom{12}{4} = 495$ and $\binom{11}{4} = 330$, so the final answer is $825 \cdot 11!$.
A: From context I assume $P_i^j = \frac{j!}{(i-1)!}$, but then I get different numbers than you:
$$P_5^{15}+11P_5^{14} = 94443148800$$
and
$$P_4^{14} + 11P_4^{13} = 25945920000$$
$$11P_4^{13} + 11\cdot10\cdot P_4^{12} = 20197900800$$
$$11\cdot 10\cdot P_4^{12} + 11\cdot 10\cdot 9\cdot P_3^{11} = 15367968000$$
and, at least, the second group sums to less than the first.
