binom{15}{8} + binom{15}{9} - binom{15}{6} - | vedclass

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(A) Using the Pascal's identity,$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\left( \binom{15}{8} + \binom{15}{9} \right) - \le

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(A) Using the Pascal's identity,$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\left( \binom{15}{8} + \binom{15}{9} \right) - \left( \binom{15}{6} + \binom{15}{7} \right)$.Applying the identity:$\binom{15}{8} + \binom{15}{9} = \binom{16}{9}$.$\binom{15}{6} + \binom{15}{7} = \binom{16}{7}$.So,the expression becomes $\binom{16}{9} - \binom{16}{7}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,we have $\binom{16}{7} = \binom{16}{16-7} = \binom{16}{9}$.Therefore,$\binom{16}{9} - \binom{16}{9} = 0$.

$\binom{15}{8} + \binom{15}{9} - \binom{15}{6} - \binom{15}{7} = \dots$

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