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It happens that I have to use combinatorial and I am trying to study by my own this interesting subject. I have two questions on which I am stuck. Any help will be appreciated.

Let $n_i\in \{0, 1, \dots, 2n\}$ for $i=1, \dots, 2p$ be such that $n_1+...+n_{2p}=2n.$ We have $k= $ $2n+2p-1 \choose 2n$ ways to choose $n_i$ so that $n_1+\dots+n_{2p}=2n$, so we have $k$ $2p$-tuples $(n_1^j,\dots, n_{2p}^j)$, $j=1, \dots ,k$. Let $M^j$ be the number of the odd elements $n_i$ in the $j$ -th tuple.

I have two questions:

1). Is it possible to write in a more compact form the following formula:

$$\begin{align*}&\left[\binom{2p-M}p+\binom{M}2\binom{2p-M}{p-2}+\binom{M}4\binom{2p-M}{p-4}+\dots+\binom{M}p\right]\\ &\qquad\qquad-\left[\binom{M}1\binom{2p-M}{p-1}+\binom{M}3\binom{2p-M}{p-3}+\dots+\binom{M}{M-1}\binom{2p-M}1\right]\\\\ &=C_1(p,M)-C_2(p,M) \end{align*}\;,$$

where $C_1(p,M)$ and $C_2(p,M)$ are the terms in square brackets above.

2). For each $M^j$, $j=1,...,k$, is it possible to find $$\max_{j}(C_1(p,M^j)-C_2(p,M^j))\;?$$

I've tried to use Vandermonde's identity for the first question, but it did not work. Thank you.

share|cite|improve this question
What are $C_M$ and $C_{2p-M}$? Catalan numbers? And are $C_1(p,M)$ and $C_2(p,M)$ defined by the two sums in the long expression? – Brian M. Scott Mar 1 '12 at 4:24
Yes. $C_1(p, M) $ and $C_2(p,M)$ defined by the two long sums. By $C_n^m$ defined expression $C_n^m=\frac{n!}{m!(n-m)!}$ – Michael Mar 1 '12 at 4:47
Okay: to make it easier to read, I’m going to replace $C_n^m$ by $\binom{n}m$. – Brian M. Scott Mar 1 '12 at 4:54

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