18 boys and 2 girls are made to stand in a random order.Let X be the number of boys standing in between two girls . 18 boys and 2 girls are made to stand in a random order.Let X be the number of boys standing in between two girls . 
Find $ P(X=5).$
Initially i thought that the distribution is uniform , but unfortunately it is not.
Here is what i did :-
Ways to select 5 boys from 18 $=$ $18\choose 5$
The denominator then becomes=$\displaystyle\sum_{18}^{i=0}$$ 18\choose i $$=2^{18}\Rightarrow P(X=5)=\frac{18\choose 5}{2^{18}}$
But the answer is $\frac{14}{190}$ .How do i get this answer (if it is correct!)??
 A: The denominator is ${20 \choose 2} = 190$: there are $20$ total boys and girls, and if you number the positions $1$ through $20$, then there are two positions out of those $20$ total for the two girls to stand.
The numerator is $14$: there are $14$ ways for two girls to have $5$ boys in between: they can have positions $1$ and $7$, or $2$ and $8$ ... all the way to $14$ and $20$
A: Here is what is wrong with your method. I assume you are using the following method: to find the probability of an event $E$ in a sample space $S$ where every outcome is equally likely, the answer is $|E|/|S|$. A common mistake is choosing the probability space incorrectly so that not all outcomes are equally likely.
In your denominator, you have $2^{18}$, implying that your probability space $S$ consists of all $2^{18}$ choices for the subset of boys to go between the girls. The problem is that not all of these choices are equally likely, so that $|E|/|S|$ method does not work. 
In this problem, we assume that the boys and girls are standing in a random order. This means we should assume all $20!$ orderings of the people are equally likely, so these orderings should be our sample space, $S$. (There are other valid choices, as in Bram28's answer). Therefore, we must count orderings where there are 5 boys between the girls. This is more information than you calculated; you only specified which boys go in the middle.
