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

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(C) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.Expanding the summation: $^{47}C_4 + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{

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$^{52}C_4$

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(C) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.Expanding the summation: $^{47}C_4 + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3)$.Using the identity $^{47}C_4 + ^{47}C_3 = ^{48}C_4$,the expression becomes: $^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{48}C_4$.Applying the identity again: $^{48}C_3 + ^{48}C_4 = ^{49}C_4$,so we have: $^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{49}C_4$.Continuing this process: $^{49}C_3 + ^{49}C_4 = ^{50}C_4$,so we have: $^{51}C_3 + ^{50}C_3 + ^{50}C_4$.Then: $^{50}C_3 + ^{50}C_4 = ^{51}C_4$,so we have: $^{51}C_3 + ^{51}C_4$.Finally: $^{51}C_3 + ^{51}C_4 = ^{52}C_4$.

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

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