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

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

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$^{56}{C_4}$

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(B) The given expression is ${}^{50}{C_4} + \sum_{r = 1}^6 {^{56 - r}{C_3}}$.Expanding the summation,we get: ${}^{50}{C_4} + {}^{55}{C_3} + {}^{54}{C_3} + {}^{53}{C_3} + {}^{52}{C_3} + {}^{51}{C_3} + {}^{50}{C_3}$.Rearranging the terms: ${}^{50}{C_4} + {}^{50}{C_3} + {}^{51}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$.Using the Pascal's identity ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$,we solve step by step:$({}^{50}{C_4} + {}^{50}{C_3}) + {}^{51}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3} = {}^{51}{C_4} + {}^{51}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$.$({}^{51}{C_4} + {}^{51}{C_3}) + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3} = {}^{52}{C_4} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}$.Continuing this process,we get ${}^{53}{C_4} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3} = {}^{54}{C_4} + {}^{54}{C_3} + {}^{55}{C_3} = {}^{55}{C_4} + {}^{55}{C_3} = {}^{56}{C_4}$.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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