binom{50}{4} + sum_{i=1}^{6} binom{56-i}{3} = | vedclass

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(D) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \binom{50

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$\binom{56}{4}$

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(D) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \binom{50}{4} + \binom{55}{3} + \binom{54}{3} + \binom{53}{3} + \binom{52}{3} + \binom{51}{3} + \binom{50}{3}$.Rearranging the terms:$= \binom{50}{4} + \binom{50}{3} + \binom{51}{3} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.Using $\binom{50}{4} + \binom{50}{3} = \binom{51}{4}$:$= \binom{51}{4} + \binom{51}{3} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.Using $\binom{51}{4} + \binom{51}{3} = \binom{52}{4}$:$= \binom{52}{4} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.Continuing this process:$= \binom{53}{4} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3} = \binom{54}{4} + \binom{54}{3} + \binom{55}{3}$.$= \binom{55}{4} + \binom{55}{3} = \binom{56}{4}$.

$\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \dots$

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