Question

${}^{47}{C}_4+{\sum }_{j=1}^5{}^{52-j}{C}_3=$

• A. ${}^{52}{C}_4$
• B. ${}^{51}{C}_4$
• C. 9${}^{52}{C}_3$
• D. None of these

Answer 1

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

We have,

${}^{47}{C}_4+\sum _{j=1}^5{}^{52-j}{C}_3{=}^{47}{C}_4{+}^{51}{C}_3{+}^{50}{C}_3{+}^{49}{C}_3{+}^{48}{C}_3{+}^{47}{C}_3$

$=\left({}^{47}{C_4+}^{47}{C}_3\right){+}^{48}{C}_3{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3$

${=}^{48}{C_4+}^{48}{C}_3{+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3$

${=}^{49}{C_4+}^{49}{C}_3{+}^{50}{C}_3{+}^{51}{C}_3{=}^{50}{C}_4{+}^{50}{C}_3{+}^{51}{C}_3$

${=}^{51}{C}_4{+}^{51}{C}_3{=}^{52}{C}_4.$