Question

Find the value of ${}^{47}{C}_4+\sum _{r=1}^5$ ${}^{52-r}{C}_3=?$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

${}^{47}{C}_4+\sum _{r=1}^5$${}^{52-r}{C}_3{=}^{47}{C}_4{+}^{51}{C}_3{+}^{50}{C}_3{+}^{49}{C}_3{+}^{48}{C}_3{+}^{47}{C}_3$
We know,
${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$
Simplify it by taking two terms at a time.
${}^{47}{C}_4+\sum _{r=1}^5$${}^{52-r}{C}_3{=}^{47}{C}_4{+}^{47}{C}_3{+}^{48}{C}_3{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3$
=${}^{48}{C}_4{+}^{48}{C}_3{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3$
=${}^{49}{C}_4{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3$
=${}^{50}{C}_4{+}^{50}{C}_3{+}^{51}{C}_3$
=${}^{51}{C}_4{+}^{51}{C}_3$
=${}^{52}{C}_4$