Can one simplify this expression involving products of binomial coefficients?

I was wondering if there is a way to simplify the following expression. $$N, M, L, K, n, Q$$ are fixed natural numbers. Also n is supposed to be even.

\begin{align*} \sum_{q=0}^Q & \sum_{l=0}^L (-1)^{Q-q+l} \binom{N}{q} \binom{M}{Q-q} \binom{L}{l} \binom{q}{l} \cdot \\ \cdot & \binom{Q-q}{L-l+K} \frac{l! (L-l+K)!}{(l+n/2-1)!(L-l+K+n/2-1)!} \end{align*}

Hope someone can help me.

• what are the three dots meaning ? – G Cab Sep 14 '20 at 15:43
• Just wanted to make it look better... – gameranger Sep 14 '20 at 16:39
• I mean, is it just a continuation of the multiplication of the shown terms , is it? then please use just one dot – G Cab Sep 14 '20 at 16:45
• Thanks, edited it – gameranger Sep 14 '20 at 18:44

Since the expression is complicate, let's divide the summand into two blocks \eqalign{ & A(q,l) = \left( { - 1} \right)^{Q - q + l} \left( \matrix{ N \cr q \cr} \right) \left( \matrix{ q \cr l \cr} \right)\left( \matrix{ M \cr Q - q \cr} \right) \left( \matrix{ Q - q \cr L + K - l \cr} \right) \cr & B(l) = \left( \matrix{ L \cr l \cr} \right) {{l!\left( {L + K - l} \right)!} \over {\left( {l + n/2 - 1} \right)!\left( {L + K - l + n/2 - 1} \right)!}} \cr} and treat them separately at first.

a) Reduction of the A block

Let's consider the four binomials depending on $$q$$ \eqalign{ & A(q,l) = \left( { - 1} \right)^{Q - q + l} \left( \matrix{ N \cr q \cr} \right) \left( \matrix{ q \cr l \cr} \right)\left( \matrix{ M \cr Q - q \cr} \right) \left( \matrix{ Q - q \cr L + K - l \cr} \right) = \cr & = \left( { - 1} \right)^{Q - q + l} \left( \matrix{ N \cr l \cr} \right) \left( \matrix{ N - l \cr q - l \cr} \right)\left( \matrix{ M \cr L + K - l \cr} \right) \left( \matrix{ M - L - K + l \cr Q - q - L - K + l \cr} \right) = \cr & = \left( { - 1} \right)^{Q - q + l} \left( \matrix{ N - l \cr q - l \cr} \right) \left( \matrix{ M - L - K + l \cr Q - L - K - \left( {q - l} \right) \cr} \right) \left( \matrix{ N \cr l \cr} \right)\left( \matrix{ M \cr L + K - l \cr} \right) \cr & = A_{\,1} (q,l) \, A_{\,2} (l) \cr} where we have applied the Trinomial Revision to both blocks, and rearranged them:
now we have only two binomials depending on $$q$$.

b) Separation of the sum in $$q$$

We can omit the upper bound on the two sums, since they are implicit in the binomials and write \eqalign{ & S = \sum\limits_{q = 0}^Q {\sum\limits_{l = 0}^L {A_{\,1} (q,l)\,A_{\,2} (l)\,B(l)} } = \sum\limits_{0\, \le \,q} {\sum\limits_{0\, \le \,l} {A_{\,1} (q,l)\,A_{\,2} (l)\,B(l)} } = \cr & = \sum\limits_{0\, \le \,l} {\left( {\sum\limits_{0\, \le \,q} {A_{\,1} (q,l)\,} } \right)A_{\,2} (l)\,B(l)} \cr}

The internal sum is \eqalign{ & S_{\,1} (l) = \sum\limits_{0\, \le \,q} {A_{\,1} (q,l)\,} = \cr & = \sum\limits_{0\, \le \,q} {\left( { - 1} \right)^{Q - q + l} \left( \matrix{ N - l \cr q - l \cr} \right) \left( \matrix{ M - L - K + l \cr Q - L - K - \left( {q - l} \right) \cr} \right)\,} = \cr & = \sum\limits_{0\, \le \,j\,\,\left( { \le \,Q - L - K} \right)} { \left( { - 1} \right)^{Q - j} \left( \matrix{ N - l \cr j \cr} \right) \left( \matrix{ M - L - K + l \cr Q - L - K - j \cr} \right)\,} \cr} and - for general values of the parameters - the presence of $$(-1)^j$$ does not allow to put it in a closed form, if not in terms of a hypergeometric function, containing $$l$$ in its parameters, and computed at $$z = -1$$. The advantage of a hypergeometric expression just depends on the use you are going to do of the whole expression.

c) The terms in $$l$$

The terms in $$l$$, besides $$S_1(l)$$, can be recasted in many different ways but - again in general - there will be no any significant simplification.
Also here one possible expression would be through a Generalized Hypergeometric with many parameters.