Question

The value of ${}^{50}C_4+\sum _{r=1}^6\left({}^{56-r}C_3\right)$ is

• A. ${}^{56}C_4$
• B. ${}^{56}C_3$
• C. ${}^{55}C_3$
• D. ${}^{55}C_4$

Answer 1

The correct option is A

${}^{56}C_4$

Explanation for the correct option:

Given expression, ${}^{50}C_4+\sum _{r=1}^6\left({}^{56-r}C_3\right)$ can be expanded as,

${}^{50}C_4+{}^{55}C_3+{}^{54}C_3+{}^{53}C_3+{}^{52}C_3+{}^{51}C_3+{}^{50}C_3$

By the formula,${}^{n}C_r+{}^{n}C_{r-1}={}^{n+1}C_r$

Hence , let us arrange the order first,

$$\begin{gathered} {}^{50}C_4+{}^{50}C_3+{}^{51}C_3+{}^{52}C_3+{}^{53}C_3+{}^{54}C_3+{}^{55}C_3 \\ ={}^{51}C_4+{}^{51}C_3+{}^{52}C_3+{}^{53}C_3+{}^{54}C_3+{}^{55}C_3 \\ ={}^{52}C_4+{}^{52}C_3+{}^{53}C_3+{}^{54}C_3+{}^{55}C_3 \\ ={}^{53}C_4+{}^{53}C_3+{}^{54}C_3+{}^{55}C_3 \\ ={}^{54}C_4+{}^{54}C_3+{}^{55}C_3 \\ ={}^{55}C_4+{}^{55}C_3 \\ ={}^{56}C_4 \end{gathered}$$

Hence, the correct answer is option (A), ${}^{56}C_4$.