The value of {}^{47}C_4 + sum_{j=1}^5 {}^{(52 | vedclass

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

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${}^{52}C_4$

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(A) We use the identity ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$. 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 the identity ${}^{47}C_4 + {}^{47}C_3 = {}^{48}C_4$: $S = ({}^{48}C_4 + {}^{48}C_3) + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3$. Using the identity ${}^{48}C_4 + {}^{48}C_3 = {}^{49}C_4$: $S = ({}^{49}C_4 + {}^{49}C_3) + {}^{50}C_3 + {}^{51}C_3$. Continuing this process: $S = ({}^{50}C_4 + {}^{50}C_3) + {}^{51}C_3 = {}^{51}C_4 + {}^{51}C_3 = {}^{52}C_4$. Thus,the correct option is $A$.

The value of ${}^{47}C_4 + \sum_{j=1}^5 {}^{(52-j)}C_3$ is

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