<s>[INST] An identity involving a product of two binomial coefficients I'm trying to find a closed formula (in the parameters $q,N$) for the following sum:
$$ \sum_{k=q}^{N} {{k-1}\choose{q-1}} {{k}\choose {q}} $$
Can anybody give me a lead? 
Lior 
 [/INST] This is a hypergeometric sum (the ratio of two consecutive terms is a rational function) and is therefore susceptible to Gosper-type algorithms, of which you can read more in the book A = B.
I believe Mathematica has got Gosper's algorithm implemented. It says:
In[38]:= s[n_, q_] := Sum[Binomial[k - 1, q - 1] * Binomial[k, q], {k, q, n}]

In[40]:= FullSimplify[s[n, q]]

Out[40]= Gamma[-2 q]/(Gamma[1 - q] Gamma[-q]) - Binomial[n, -1 + q] Binomial[1 + n, 
q] HypergeometricPFQ[{1, 1 + n, 2 + n}, {2 + n - q, 2 + n - q}, 1]

The answer is that "this is some hypergeomteric series". This makes me think there is no closed form solution. We can plug in specific values of $q$ in which case we do get specific closed forms:
\begin{array}{l|l}
 q & \text{closed form} \\ \hline
 1 & \frac{n^2}{2}+\frac{n}{2} \\
 2 & \frac{n^4}{8}-\frac{n^3}{12}-\frac{n^2}{8}+\frac{n}{12} \\
 3 & \frac{n^6}{72}-\frac{7 n^5}{120}+\frac{n^4}{18}+\frac{n^3}{24}-\frac{5
   n^2}{72}+\frac{n}{60} \\
 4 & \frac{n^8}{1152}-\frac{17 n^7}{2016}+\frac{17 n^6}{576}-\frac{29
   n^5}{720}+\frac{n^4}{1152}+\frac{13 n^3}{288}-\frac{n^2}{32}+\frac{n}{280} \\
 5 & \frac{n^{10}}{28800}-\frac{31 n^9}{51840}+\frac{n^8}{240}-\frac{179
   n^7}{12096}+\frac{773 n^6}{28800}-\frac{301 n^5}{17280}-\frac{47 n^4}{2880}+\frac{83
   n^3}{2592}-\frac{53 n^2}{3600}+\frac{n}{1260} \\
\end{array}
I am not a combinatorialist, so these numbers seem completely arbitrary to me.
</s>