13 boys and 2 girls are to be placed next to each other. What is the probability that there are exactly 4 boys between the 2 girls? $13$ boys and $2$ girls are to be placed next to each. What is the probability that there are exactly $4$ boys between the $2$ girls?
Arrangements possible:

*

*$GBBBBGBBBBBBBBB$

*$BGBBBBGBBBBBBBB$

*$BBGBBBBGBBBBBBB$

*$BBBGBBBBGBBBBBB$

*$BBBBGBBBBGBBBBB$

*$BBBBBGBBBBGBBBB$

*$BBBBBBGBBBBGBBB$

*$BBBBBBBBGBBBBGB$

*$BBBBBBBBBGBBBBG$
So the number of ways the arrangement can happen is $C(13,9) = 715$
I need help beyond this to frame the solution.
 A: You overlooked BBBBBBBGBBBBGBB.

If we only discern on boy/girl then we must find the number of tuples $(a,b)$ where $a$ and $b$ are non-negative integers with $a+b=9$. With stars and bars (string GBBBBG serves as bar) we find $\binom{9+1}1=10$ solutions (not $9$). Here $a$ corresponds with the number of boys on left and $b$ with the number of boys on right.
This must be multiplied with $2!\times13!$ if we discern persons, so the number of favourable arrangements is:$$10\times2!\times13!$$
The number of possible arrangements is $15!$ so we end up with probability:$$\frac{10\times2!\times13!}{15!}=\frac{2}{21}$$
A: Use Laplace's rule: $$P(A) = \frac{\mbox{Favorable cases}}{\mbox{Total Cases}}$$
The total number of cases in this problem is how you can arrange 15 people, which is $15!$. Now let's analyse the favorable cases.
First of all, you have to choose where to place one of the girls in the 15 places, since the place of the other girl will automatically be determined 4 seats away. Then, we choose the seat: $\displaystyle \binom{10}{1}$. We can't choose any seat further than the tenth, since there are no solutions further away. We can place any of the two girls, so we multiply the result by two.
Now we have to arrange the boys in the other 13 places left, which can be done in $13!$ ways. Summing up, if we call $A$ the event of having 2 girls separated by 4 boys,
$$P(A) = \frac{\displaystyle \binom{10}{1} \cdot 2 \cdot 13!}{15!} = \frac{2}{21}$$
A: You can generate a placement by first choosing a first position for a girl uniformly at random from the $15$ available positions, then from the remaining $14$ positions choose another position for a girl. Even though those $15\times14$ possibilities produce each possible outcome twice (for two permutations of the $2$ chosen positions), it still chooses a final outcome uniformly at random (precisely because every possible outcome is produced in the same number of ways). Now the event you are interested in occurs if the second choice is either $5$ positions to the left, or $5$ positions to the right of the first choice. Given a first choice this is always possible, but not in the same number of ways: the first $5$ choices from either end do not allow a second choice at distance $5$ in that direction, so they only allow a second choice at the opposite side. So the number of possibilities at the second choice depends on the first choice as follows: $1,1,1,1,1,2,2,2,2,2,1,1,1,1,1$. So given the first choice the probability of a favourable second choice is either $\frac1{14}$ or $\frac2{14}$, according to this table. Since all first choices are equally likely, our answer will be the average over these $15$ values: $(10\times\frac1{14}+5\times\frac2{14})/15=\frac{20}{210}=\frac2{21}$.
A: Consider a "valid" block of  $2$ girls, $4$ boys $\boxed{GBBBBG}$
It can be inserted in $10$ ways among the $9$ remaining boys
$\uparrow B\uparrow B\uparrow B\uparrow B\uparrow B\uparrow B \uparrow B\uparrow B \uparrow  B \uparrow $
Permuting the girls and boys in each such configuration, we get
$Pr = \dfrac{10\times 2!13!}{15!} = \dfrac2{21}$
A: It does not matter which boy is which, nor which girl is which. The only thing that matters is the locations of the two girls within the row.
There are $\binom{15}{2}=\frac{15\cdot 14}{2} = 105$ pairs of locations that the girls could end up in, all of which are equally likely.
Only $10$ possible arrangements with $4$ seats between the two girls. With $4$ boys between the girls, the remaining $9$ boys can be distributed in $10$ ways (any number $0$ to $9$ to the left of the left-most girl, the rest to the right of the right-most girl).
The probability of picking one of these $10$ arrangements out of the $105$ possibilities is $\frac{10}{105}=\frac{2}{21}$.
A: Choose boys: $\displaystyle\binom{13}{4}$
Call the ‘GBBBBG’ as a unit U. Thus now there are 13-4+1=10 entities: BBBBBBBBBU to be arranged. Arrange in 10! ways.
Now, the four boys in U are arranged in 4! ways and the girls in 2! ways. Thus, final answer is $$\binom {13}{4}\cdot 10!4!2!.$$$$=\frac{13!}{4!9!}\cdot 10!4!2!=20\times 13!$$
So the probability is $$\frac{20\times 13!}{15!}=\frac{20}{14\times 15}=\frac{2}{21}$$
A: For those who dislike doing any extra arithmetic:
Step 1: Find number of favorable arrangements. Arrange all the boys in a row, bbbbbbbbbbbbb, there are B ways to do this. Place the girls so that they are 4 boys apart, as your example shows there are 20 ways to do this (10*2 because either girl could be first). So there are 20*B favorable arrangements.
Step 2: Find total arrangements. Arrange the boys as previously, bbbbbbbbbbbbb, there are still B ways to do this. Place the first girl anywhere in the line, there are 14 ways to do this. Place the second girl anywhere into the line, there are 15 ways to do this after the first girl has been placed. So there are 14*15*B total arrangements.
Step 3: Divide, favorable/total: 20*B/(14*15*B), B cancels, a factor of 2 and a factor of 5 cancel, final answer 2/21.
