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
Solution 1:
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.