Question

${}^{n-2}C_r+2{}^{n-2}C_{r-1}+{}^{n-2}C_{r-2}$ is equal to

• A. ${}^{n+1}C_r$
• B. ${}^{n}C_r$
• C. ${}^{n+1}C_{r+1}$
• D. ${}^{n-1}C_r$

Answer 1

The correct option is B

${}^{n}C_r$

Explanation for the correct answer:

Calculating the value of the given expression:

Given Expression is ${}^{n-2}C_r+2{}^{n-2}C_{r-1}+{}^{n-2}C_{r-2}$

Now, expanding the middle term we get the expression as

$\left({}^{n-2}C_r+{}^{n-2}C_{r-1}\right)+\left({}^{n-2}C_{r-1}+{}^{n-2}C_{r-2}\right)$

Using the property ${}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{=}{}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{C}_{\mathbf{r}}$ in both places under brackets we get,

$$\begin{gathered} =\left({}^{n-2+1}C_r\right)+\left({}^{n-2+1}C_{r-1}\right) \\ =\left({}^{n-1}C_r\right)+\left({}^{n-1}C_{r-1}\right) \end{gathered}$$

Again using the same property we get,

$={}^{n}C_r\left[\because {}^{n-1+1}C_r\right]$

Hence, option (B) is the correct answer.