The value of {}^{50}C_4 + sum_{r=1}^{6} {}^{5 | vedclass

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(B) We are given the expression $S = {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$. Expanding the summation,we get $S = {}^{50}C_4 + ({}^{55}C_3 + {}^{54}

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${}^{56}C_4$

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(B) We are given the expression $S = {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$. Expanding the summation,we get $S = {}^{50}C_4 + ({}^{55}C_3 + {}^{54}C_3 + {}^{53}C_3 + {}^{52}C_3 + {}^{51}C_3 + {}^{50}C_3)$. Rearranging the terms,$S = ({}^{50}C_4 + {}^{50}C_3) + {}^{51}C_3 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$. Using the Pascal's identity ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$,we have ${}^{50}C_4 + {}^{50}C_3 = {}^{51}C_4$. Now,$S = ({}^{51}C_4 + {}^{51}C_3) + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{52}C_4 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$. Continuing this process,$S = ({}^{52}C_4 + {}^{52}C_3) + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{53}C_4 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$. $S = ({}^{53}C_4 + {}^{53}C_3) + {}^{54}C_3 + {}^{55}C_3 = {}^{54}C_4 + {}^{54}C_3 + {}^{55}C_3$. $S = ({}^{54}C_4 + {}^{54}C_3) + {}^{55}C_3 = {}^{55}C_4 + {}^{55}C_3 = {}^{56}C_4$. Thus,the correct option is $B$.

The value of ${}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$ is

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