<s>[INST] Summation of a binomial series Sum of the series $\left( {\begin{array}{*{20}{c}}
{20}\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{20}\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
6
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{20}\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
7
\end{array}} \right) - .... - \left( {\begin{array}{*{20}{c}}
{20}\\
{15}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
{20}
\end{array}} \right)$ is____
My approach is as follow
$T = \left( {\begin{array}{*{20}{c}}
{20}\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{20}\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
6
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{20}\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
7
\end{array}} \right) - .... - \left( {\begin{array}{*{20}{c}}
{20}\\
{15}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
{20}
\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}
n\\
p
\end{array}} \right) = {}^n{C_p}$
$T = \sum\limits_{r = 0}^{15} {{{\left( { - 1} \right)}^r}.{}^{20}{C_r}.{}^{20}{C_{r + 5}}}  \Rightarrow T = \sum\limits_{r = 1}^{15} {{{\left( { - 1} \right)}^r}.\frac{{20!}}{{r!\left( {20 - r} \right)!}}.\frac{{20!}}{{\left( {r + 5} \right)!\left( {15 - r} \right)!}}}  \Rightarrow T = \frac{{20!}}{{15!}} \times \frac{{20!}}{{25!}}\sum\limits_{r = 0}^{15} {{{\left( { - 1} \right)}^r}.\frac{{15!}}{{r!\left( {15 - r} \right)!}}.\frac{{25!}}{{\left( {r + 5} \right)!\left( {20 - r} \right)!}}} $
How do we approach from here
 [/INST] Hint :
Consider the product $(x-1)^{20}(1+x)^{20}$. Verify by expansion and multiplication that coefficient of $x^{25}$ is
$$\binom{20}{0}\binom{20}{5} - \binom{20}{1}\binom{20}{6}+\ldots - \binom{20}{15}\binom{20}{20}$$
Now find the coefficient of $x^{25}$ in $(x-1)^{20}(1+x)^{20}$ using a different way.  The two must be equal.
</s>