The value of sumlimits_{r = 0}^{15} {left( {{ | vedclass

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(A) We use the identity ${}^{n}{C_r} = {}^{n}{C_{n-r}}$.The expression is $\sum\limits_{r = 0}^{15} {\left( {}^{15}{C_r} \cdot {}^{40}{C_{15}} \cdot {

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(A) We use the identity ${}^{n}{C_r} = {}^{n}{C_{n-r}}$.The expression is $\sum\limits_{r = 0}^{15} {\left( {}^{15}{C_r} \cdot {}^{40}{C_{15}} \cdot {}^{20}{C_r} - {}^{35}{C_{15}} \cdot {}^{15}{C_r} \cdot {}^{25}{C_r} \right)}$.Rewrite the terms using the identity ${}^{20}{C_r} = {}^{20}{C_{20-r}}$ and ${}^{25}{C_r} = {}^{25}{C_{25-r}}$:$= {}^{40}{C_{15}} \sum\limits_{r = 0}^{15} {({}^{15}{C_r} \cdot {}^{20}{C_{20-r}})} - {}^{35}{C_{15}} \sum\limits_{r = 0}^{15} {({}^{15}{C_r} \cdot {}^{25}{C_{25-r}})}$.Using Vandermonde's Identity,$\sum\limits_{k=0}^{r} {}^{m}{C_k} \cdot {}^{n}{C_{r-k}} = {}^{m+n}{C_r}$:$= {}^{40}{C_{15}} \cdot {}^{35}{C_{20}} - {}^{35}{C_{15}} \cdot {}^{40}{C_{25}}$.Since ${}^{35}{C_{20}} = {}^{35}{C_{15}}$ and ${}^{40}{C_{25}} = {}^{40}{C_{15}}$:$= {}^{40}{C_{15}} \cdot {}^{35}{C_{15}} - {}^{35}{C_{15}} \cdot {}^{40}{C_{15}} = 0$.

The value of $\sum\limits_{r = 0}^{15} {\left( {{}^{15}{C_r} \cdot {}^{40}{C_{15}} \cdot {}^{20}{C_r} - {}^{35}{C_{15}} \cdot {}^{15}{C_r} \cdot {}^{25}{C_r}} \right)}$ is:

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