Question

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

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

Answer 1

The correct option is C ${}^{52}{C}_4$

$Here{,}^{47}{C}_4+{\sum }_{j=1}^5{}^{52-j}{C}_3{=}^{47}{C}_4{+}^{51}{C}_3{+}^{50}{C}_3{+}^{49}{C}_3{+}^{48}{C}_3{+}^{47}{C}_3={(}^{47}{C}_4{+}^{47}{C}_3){+}^{48}{C}_3{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3$
$\left[Using{}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r\right]$
$={(}^{48}{C}_4{+}^{48}{C}_3){+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3={(}^{49}{C}_4{+}^{49}{C}_3){+}^{50}{C}_3{+}^{51}{C}_3={(}^{50}{C}_4{+}^{50}{C}_3){+}^{50}{C}_3{+}^{51}{C}_3={(}^{51}{C}_4{+}^{51}{C}_3){=}^{52}{C}_4$