For the mean
\begin{align*}
\sum_{x=0}^n x \frac{\binom{n_1}{x} \binom{n-n_1}{r-x}}{\binom{n}{r}} & =
\frac{1}{\binom{n}{r}} \sum_{x=1}^n \frac{n_1(n_1-1)!}{(x-1)!(n_1-x)!} \binom{n-n_1}{r-x}\\
& = \frac{n_1}{\binom{n}{r}} \sum_{x=1}^n \frac{(n_1-1)!}{(x-1)!((n_1-1)-(x-1))!} \binom{n-n_1}{r-x} \\
& = \frac{n_1}{\frac{n(n-1)!}{r!(n-r)!}} \sum_{x=1}^n \binom{n_1-1}{x-1} \binom{n-n_1}{r-x}
\end{align*}
Consider the following change of variables for the summation:
\begin{gather*}
y = x-1\\
n_3 = n_1-1\\
s = r-1\\
m = n-1
\end{gather*}
Then, this becomes
\begin{align*}
\mu = \sum_{x=0}^n x \frac{\binom{n_1}{x} \binom{n-n_1}{r-x}}{\binom{n}{r}} & = r \frac{n_1}{n} \sum_{y=0}^m \frac{\binom{n_3}{y} \binom{m-n_3}{s-y}}{\binom{m}{s}}\\
& = r \frac{n_1}{n} \cdot 1
\end{align*}
noting that the summation is in the same form as was show yields 1 above.