I am trying to solve this problem but I am not sure how to obtain the formula given below. Any help would be appreciated.
A boy is selected at random from among the children belonging to families with $n$ children. It is known that the boy has at least $2$ sisters. Show that the probability that he has $k-1$ brothers is:
$$\frac {(n-1)!}{(2^{n-1}-n)(k-1)!(n-k)!}$$
Let's let $B$ be the number of brothers and $S$ be the number of sisters. Then we want to find $P(B = k-1 \mid S \geq 2)$.
$$P(B = k-1 \mid S \geq 2) = \frac{P(B = k-1 \cap S \geq 2)}{P(S \geq 2)} = \frac{P(S \geq 2 \mid B = k-1)P(B = k-1)}{P(S \geq 2)},$$
which is from Bayes.
Now, since there are $n-1$ children besides the boy in question, for the probability that $B = k-1$ we have the binomial probability $$P(B = k-1) = \binom{n-1}{k-1}\cdot \frac{1}{2^{n-1}} = \frac{(n-1)!}{(k-1)!(n-k)!\left(2^{n-1}\right)},$$ assuming genders are equally likely among remaining children.
For the denominator, the probability of at least two sisters we have
$$P(S \geq 2) = 1 - P(S = 0) - P(S = 1) = 1 - \frac{1}{2^{n-1}} - \frac{n-1}{2^{n-1}} = \frac{2^{n-1}-n}{2^{n-1}}.$$
So now we're already getting pretty close to the expression in your question. Can you take it from here?
