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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.

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

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