Question

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

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

Answer 1

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

We have,

$\begin{array}{c}{}^{47}{C}_4{+}^{51}{C}_3{+}^{50}{C}_3{+}^{49}{C}_3{+}^{48}{C}_3{+}^{47}{C}_3\\ {=}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r\\ {=}^{48}{C}_4{+}^{48}{C}_3{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3\\ {=}^{49}{C}_4{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3\end{array}$

Similarly,

${=}^{52}{C}_4$.