Question

The expression ${}^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}$ is equal to

• A. ${}^{n+4}{C}_r$
• B. $2{·}^{n+4}{C}_{r-1}$
• C. $4{·}^n{C}_r$
• D. $11{·}^n{C}_r$

Answer 1

The correct option is A

${}^{n+4}{C}_r$

Explanation for the correct answer:

Evaluate the value of the given expression.

The expression ${}^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}$ is given.

Simplify the given expression as follows.

${}^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}=\left({}^n{C}_r{+}^n{C}_{r-1}\right)+3·\left({}^n{C}_{r-1}{+}^n{C}_{r-2}\right)+3·\left({}^n{C}_{r-2}{+}^n{C}_{r-3}\right)+\left({}^n{C}_{r-3}{+}^n{C}_{r-4}\right)$

We know that, ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$

Therefore,

$$\begin{gathered} {}^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}{=}^{n+1}{C}_r+3{·}^{n+1}{C}_{r-1}+3{·}^{n+1}{C}_{r-2}{+}^{n+1}{C}_{r-3} \\ {\Rightarrow }^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}=\left({}^{n+1}{C}_r{+}^{n+1}{C}_{r-1}\right)+2\left({}^{n+1}{C}_{r-1}{+}^{n+1}{C}_{r-2}\right)+\left({}^{n+1}{C}_{r-2}{+}^{n+1}{C}_{r-3}\right) \\ {\Rightarrow }^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}{=}^{n+2}{C}_r+2{·}^{n+2}{C}_{r-1}{+}^{n+2}{C}_{r-2} \\ {\Rightarrow }^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}=\left({}^{n+2}{C}_r{+}^{n+2}{C}_{r-1}\right)+\left({}^{n+2}{C}_{r-1}{+}^{n+2}{C}_{r-2}\right) \\ {\Rightarrow }^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}{=}^{n+3}{C}_r{+}^{n+3}{C}_{r-1} \\ {\Rightarrow }^n{C}_r+4{·}^n{C}_{r-1}+6{·}^n{C}_{r-2}+4{·}^n{C}_{r-3}{+}^n{C}_{r-4}{=}^{n+4}{C}_r \end{gathered}$$

Therefore, the value of the given expression is equal to ${}^{n+4}{C}_r$.

Hence, option $A$ is correct .