How to calculate this triple summation?
Hint: This is just a starter which might be helpful for further calculation. We obtain a somewhat simpler representation of the triple sum.
\begin{align*} \sum_{j=1}^m&\sum_{i=j}^m\sum_{k=j}^m\frac{\binom{m}{i}\binom{m-j}{k-j}}{\binom{k}{j}}r^{k-j+i}\\ &=\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^{m}\frac{m!}{i!(m-i)!}\cdot\frac{(m-j)!}{(k-j)!(m-k)!}\cdot\frac{j!(k-j)!}{k!}r^{k-j+i}\tag{1}\\ &=\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^{m}\frac{\binom{m}{k}\binom{m}{i}}{\binom{m}{j}}r^{k-j+i}\tag{2}\\ &=\sum_{j=1}^m\frac{1}{\binom{m}{j}r^j}\sum_{i=j}^m\binom{m}{i}r^i\sum_{k=j}^m\binom{m}{k}r^k\tag{3}\\ &=\sum_{j=1}^m\frac{1}{\binom{m}{j}r^j}\left(\sum_{k=j}^m\binom{m}{k}r^k\right)^2 \end{align*}
Comment:
In (1) we write the binomial coefficients using factorials
In (2) we cancel out $(k-j)!$ and rearrange the other factorials to binomial coefficients so that each of them depends on one running index only
In (3) we can place the binomial coefficients conveniently and see that the sums with index $i$ and $k$ are the same