I have thought about the following question a long time and still got no progress. 


Currently I am writing my master thesis about association schemes in combinatorics and need an equality which seems to be super clear for everyone but me. At least they state that it's obvious.
The eigenvalues of the Johnson scheme can be derived from the Eberlein polynomials, $n$ and $k$ ($n \geq 2k$) are fixed and $i,j =0,...,k$:


\begin{equation}
 \begin{split}
 p\_i(j) 
 &= \sum\_{r = i}^{k} (-1)^{r-i+j} {r \choose i} {n - 2r \choose k - r} {n - r - j \choose r - j}\\
 &= \sum\_{r=i}^{k} (-1)^{r-i+j} {r \choose i} {k-j \choose r-j} {n-r-j \choose k-j}
 \end{split}
 \end{equation}
The second equality is just some playing around with binomial 
coefficients.


I need to show that the term above is equal to
\begin{equation}
p\_i(j) = \sum\_{r = 0}^{i} (-1)^{r} {j \choose r} {k - j \choose i - r} {n - k - j \choose i - r}.
\end{equation}
I have no idea how to do it.


If you don't know the answer in general but can help me with $j=1,2$, that'd be sufficient.


For $j=1$ the last equation is


\begin{equation}
p\_i(1) = {k - 1 \choose i} {n - k - 1 \choose i} - {k - 1 \choose i - 1} {n - k - 1 \choose i - 1}
\end{equation}


For $j=2$


\begin{equation}
p\_i(1) = {k - 2 \choose i} {n - k - 2 \choose i} - 2 {k - 2 \choose i - 1} {n - k - 2 \choose i - 1} + {k - 2 \choose i - 2} {n - k - 2 \choose i - 2}
\end{equation}


If someone knows about the polynomials or is just clever in combinatorics, I'd be super thankful. Of course I would mention you in the thesis then and spend you some beer if you come to Berlin :D



Showing equality of Eberlein polynomials
Multiply each sum by $x^i y^n$. Sum on $n$, then $i$, then $r$. In both cases we get
$$y^{k+j}(1-y)^{j-k-1}(1-x)^j(1-y+xy)^{k-j}.$$


