^{47}C_4 + sum_{r=1}^5 {}^{52-r}C_3 = | vedclass

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(C) We use the Pascal's identity: $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$.The given expression is $S = ^{47}C_4 + \sum_{r=1}^5 {}^{52-r}C_3$.Expanding

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$^{52}C_4$

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(C) We use the Pascal's identity: $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$.The given expression is $S = ^{47}C_4 + \sum_{r=1}^5 {}^{52-r}C_3$.Expanding the summation:$S = ^{47}C_4 + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3)$.Rearranging terms:$S = (^{47}C_4 + ^{47}C_3) + ^{48}C_3 + ^{49}C_3 + ^{50}C_3 + ^{51}C_3$.Using $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$:$^{47}C_4 + ^{47}C_3 = ^{48}C_4$.Now,$S = (^{48}C_4 + ^{48}C_3) + ^{49}C_3 + ^{50}C_3 + ^{51}C_3$.Using the identity again:$^{48}C_4 + ^{48}C_3 = ^{49}C_4$.Continuing this process:$S = (^{49}C_4 + ^{49}C_3) + ^{50}C_3 + ^{51}C_3 = ^{50}C_4 + ^{50}C_3 + ^{51}C_3$.$S = (^{50}C_4 + ^{50}C_3) + ^{51}C_3 = ^{51}C_4 + ^{51}C_3$.$S = ^{52}C_4$.Thus,the correct option is $C$.

$^{47}C_4 + \sum_{r=1}^5 {}^{52-r}C_3 = $

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