Question

The value of the expression ${}^{47}{C}_4+{\sum }_{f=0}^6{}^{52-f}{C}_3$ equals

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

Answer 1

The correct option is B ${}^{52}{C}_5$

Simplifying, we get
${}^{47}{C}_4+{}^{46}{C}_3+{}^{47}{C}_3+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3+{}^{52}{C}_3$

$={}^{46}{C}_3+{}^{47}{C}_4+{}^{47}{C}_3+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3+{}^{52}{C}_3$

$={}^{46}{C}_3+{}^{48}{C}_4+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3+{}^{52}{C}_3$

$={}^{46}{C}_3+{}^{49}{C}_4+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3+{}^{52}{C}_3$
:
:
:
$={}^{46}{C}_3+{}^{52}{C}_4+{}^{52}{C}_3$

$={}^{46}{C}_3+{}^{53}{C}_4$