<s>[INST] Sum of products of binomial coefficients: $\sum\limits_{k=0}^{19}\binom{18}k\binom{20}k$ Please help me find the sum given below
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}$
First I used formula $\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$  twice and got
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}=\sum_{k=0}^{19}\binom{18}{1k}\cdot \left ( \binom{19}{k}+\binom{19}{k-1} \right )=\sum_{k=0}^{19}\binom{18}{k}\cdot\left ( \binom{18}{k}+2\binom{18}{k-1}+\binom{18}{k-2} \right )$
Now have no idea what to do with that
 [/INST] 
We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}}
&=\sum_{k=0}^{18}\binom{18}{k}\binom{20}{k}\tag{1}\\
&=\sum_{k=0}^{18}\binom{18}{18-k}\binom{20}{k}\tag{2}\\
&\,\,\color{blue}{=\binom{38}{18}=33\,578\,000\,610}\tag{3}
\end{align*}

Comment:


*

*In (1) we omit the index $k=19$, since $\binom{p}{q}=0$ if $q>p$.

*In (2) we apply the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (3) we apply  Vandermonde's identity.
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