Question

The value of the expression ${}^{47}{C}_4+\sum _{j=1}^5(52-j)C_3$

• A. ${}^{47}C5$
• B. ${}^{52}{C}_5$
• C. ${}^{52}{C}_3$
• D. ${}^{49}{C}_4$

Answer 1

The correct option is C ${}^{52}{C}_3$

${}^{47}{C}_4+{\sum }_{j=1}^5(52-j)C_3$

$\Rightarrow {}^{47}{C}_4=178365$

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

$=92360$

$\Rightarrow {}^{47}{C}_4+{\sum }_{j=1}^5(52-j)C_3=178365+92360$

$=270725$

$={}^{52}{C}_3$

Hence, the answer is ${}^{52}{C}_3.$