I'm looking to find a closed form expression or to bound the following sum $$ \sum_{\atop \LARGE j\ =\ \left(k + 1\right)/3\ +\ 1} ^{\LARGE 2\left(k + 1\right)/3 \atop} {i - 1 \choose j - 1}{n - {\rm i} \choose k - j + 1} $$ where $1\leq i \leq n,\,\,0< k <n$ and $k + 1$ is divisible by $3$.
I've been able to bound it by ${n - 1\,\choose k}$, but it is not enough for the problem I'm working on.
Context: Given an input set $S$ of $n$ distinct numbers, we sample from it $k+1$ distinct numbers and get a new set $S'$. Let us call the rank of an element in a set, its position in the sorted version of the set in increasing order. For example, for the set $\{ 1, 8, 7, 3\}$, 8 would be the element of rank 4. The probability that the element of rank $i$ in the input set $S$ is the element of rank $j$ in the sampled set $S'$ is clearly $$ \frac{{i-1\choose j-1}{n-i \choose k-j+1}}{n \choose k+1}.$$ I'm interested in bounding the probability that the element of rank $i$ in the input array $S$ is in the middle third of the sampled set $S'$: $$ \frac{\sum_{\atop \LARGE j\ =\ \left(k + 1\right)/3\ +\ 1} ^{\LARGE 2\left(k + 1\right)/3 \atop} {i - 1 \choose j - 1}{n - {\rm i} \choose k - j + 1}}{n \choose k+1}$$