Question

The value of ${}^{47}{C}_4$$+\sum _{j=1}^5{}^{52-j}{C}_3$is equal to?

• A. ${}^{47}{C}_5$
• B. ${}^{52}{C}_5$
• C. ${}^{52}{C}_4$
• D. ${}^{53}C_7$

Answer 1

The correct option is C

${}^{52}{C}_4$

Explanation for the correct answer:

${}^{47}{C}_4$$+\sum _{j=1}^5{}^{52-j}{C}_3$ $={}^{47}{C}_4+{}^{51}C_3+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}C_3+{}^{47}C_3$

$={}^{47}{C}_4+{}^{47}C_3+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}C_3$

$={}^{48}{C}_4+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}C_3$ $\left(\because {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_3\right)$

$={}^{49}{C}_4+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}C_3$

$={}^{50}{C}_4+{}^{50}{C}_3+{}^{51}C_3$

$={}^{51}C_4+{}^{51}C_3$

$\Rightarrow$ ${}^{47}{C}_4$$+\sum _{j=1}^5{}^{52-j}{C}_3$$={}^{52}C_4$

Hence, the value of ${}^{47}{C}_4$$+\sum _{j=1}^5{}^{52-j}{C}_3$ is ${}^{52}{C}_4$

Hence, option $\left(C\right)$ is the correct answer.