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. ${}^{55}{C}_{33}$
• D. ${}^{55}{C}_4$

Answer 1

The correct option is A

${}^{56}{C}_4$

Explanation for correct option:
Given expression is ${}^{50}{C}_4+\sum _{r=1}^6\left({}^{56-r}{C}_3\right)$
$\Rightarrow {}^{50}{C}_4+\sum _{r=1}^6\left({}^{56-r}{C}_3\right){=}^{50}{C}_4+\left({}^{55}{C}_3{+}^{54}{C}_3{+}^{53}{C}_3{+}^{52}{C}_3{+}^{51}{C}_3{+}^{50}{C}_3\right)$
$$\begin{gathered} =\left({}^{50}{C}_4{+}^{50}{C}_3\right)+\left({}^{55}{C}_3{+}^{54}{C}_3{+}^{53}{C}_3{+}^{52}{C}_3{+}^{51}{C}_3\right) \\ {=}^{51}{C}_4{+}^{51}{C}_3+\left({}^{55}{C}_3{+}^{54}{C}_3{+}^{53}{C}_3{+}^{52}{C}_3\right)\mathbf{}\mathbf{}\mathbf{\left[}{\mathbf{\because }}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}{\mathbf{+}}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}\mathbf{-}\mathbf{1}}{\mathbf{=}}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{C}}_{\mathbf{r}}\mathbf{\right]} \\ {=}^{52}{C}_4{+}^{52}{C}_3+\left({}^{55}{C}_3{+}^{54}{C}_3{+}^{53}{C}_3\right) \\ {=}^{53}{C}_4{+}^{53}{C}_3+\left({}^{55}{C}_3{+}^{54}{C}_3\right) \\ {=}^{54}{C}_4{+}^{54}{C}_3+\left({}^{55}{C}_3\right) \\ {=}^{55}{C}_4{+}^{55}{C}_3 \\ {=}^{56}{C}_4 \end{gathered}$$

Hence, option(A) is correct