Question

# The value of $$^{47}C_4 + \displaystyle\sum_{r=1}^{5}{^{52-r}C_3}$$=

A
52C2
B
52C3
C
52C4
D
52C5

Solution

## The correct option is B $$^{52}C_4$$We know that,$$\displaystyle ^{n}{C}_{r}+{}^{n}{C}_{r-1}$$$$\displaystyle =\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$$$$\displaystyle =\frac{n!(n-r+1)}{r!(n-r)!(n-r+1)}+\frac{n!r}{(r-1)!(n-r+1)!r}$$$$\displaystyle =\frac{n!(n-r+1)}{r!(n-r+1)!}+\frac{n!r}{(n-r+1)!r!}$$$$\displaystyle =\frac{n!}{r!(n-r+1)!}(n-r+1+r)$$$$\displaystyle =\frac{n!}{r!(n-r+1)!}(n+1)$$$$\displaystyle =\frac{(n+1)!}{r!(n-r+1)!}$$$$=\displaystyle {}^{n+1}{C}_{r}$$$$\Rightarrow \displaystyle {}^{n}{C}_{r}+{}^{n}{C}_{r-1}={}^{n+1}{C}_{r}$$The equation is: $$({}^{47}{C}_{4}+{}^{47}{C}_{3})+{}^{48}{C}_{3}+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$$$=({}^{48}{C}_{4}+{}^{48}{C}_{3})+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$$$=({}^{49}{C}_{4}+{}^{49}{C}_{3})+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$$$=({}^{50}{C}_{4}+{}^{50}{C}_{3})+{}^{51}{C}_{3}$$$$=({}^{51}{C}_{4}+{}^{51}{C}_{3})$$$$={}^{52}{C}_{4}$$Hence, option 'C' is correct.Mathematics

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