binom{47}{4} + sum_{r=1}^5 binom{52-r}{3} = d | vedclass

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(C) We use the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.Given expression: $S = \binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3}$Expandi

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$\binom{52}{4}$

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(C) We use the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.Given expression: $S = \binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3}$Expanding the summation:$S = \binom{47}{4} + \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3}$Using $\binom{47}{4} + \binom{47}{3} = \binom{48}{4}$:$S = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{48}{4} = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{49}{4}$Using $\binom{49}{3} + \binom{49}{4} = \binom{50}{4}$:$S = \binom{51}{3} + \binom{50}{3} + \binom{50}{4} = \binom{51}{3} + \binom{51}{4}$Using $\binom{51}{3} + \binom{51}{4} = \binom{52}{4}$.Thus,the correct option is $C$.

$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$

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