Note: This is exact copy of my Math.SE question, which I am reposting here, as despite bounty it did not receive any answers.
Let there be $n$ pairs of shoes in a box. The the probability that from the $r \le n$ shoes I am taking out of the box there are exactly $p$ pairs is given by \begin{equation*} \mathbb{P}_{n}^{(r)}(p) = \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}}. \end{equation*} For $n = 15$ and $r \in \{6,8,10\}$. The function (assuming the continuous factorial equivalents) looks like this:
I am interested in finding the area under that curve, namely $$\int_{0}^{\frac{r}{2}} \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}} \ \text{d}p$$
I consulted this question but could derive how that would help me.
I also thought about writing the first product of binomial coefficients as $$ \binom{n}{n - p}\binom{n - p}{r - 2p} $$ which is similar to the form $\binom{f(x)}{f(y)} \binom{f(y)}{f(x)}$ mentioned in this question.