Question

The value of $\sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)$ is equal to

• A. $1124$
• B. $924$
• C. $1324$
• D. $1024$

Answer 1

The correct option is B

$924$

Explanation for the correct option:

Compute the required value.

The given expression: $\sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)$.

It is known that, ${}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}{}^{\mathbf{n}}\mathit{C}_{\mathbf{n}\mathbf{-}\mathbf{r}}$.

Thus, $\sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=\sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_r\right)$

$$\begin{gathered} \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=\sum _{r=0}^6\left[{\left({}^{6}C_r\right)}^2\right] \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)={\left({}^{6}C_0\right)}^2+{\left({}^{6}C_1\right)}^2+{\left({}^{6}C_2\right)}^2+{\left({}^{6}C_3\right)}^2+{\left({}^{6}C_4\right)}^2+{\left({}^{6}C_5\right)}^2+{\left({}^{6}C_6\right)}^2 \\ \underset{r=0}{\overset{6}{\Rightarrow \sum }}\left({}^{6}C_r·{}^{6}C_{6-r}\right)={\left({}^{6}C_0\right)}^2+{\left({}^{6}C_1\right)}^2+{\left({}^{6}C_2\right)}^2+{\left({}^{6}C_3\right)}^2+{\left({}^{6}C_{6-2}\right)}^2+{\left({}^{6}C_{6-1}\right)}^2+{\left({}^{6}C_{6-0}\right)}^2 \\ \underset{r=0}{\overset{6}{\Rightarrow \sum }}\left({}^{6}C_r·{}^{6}C_{6-r}\right)={\left({}^{6}C_0\right)}^2+{\left({}^{6}C_1\right)}^2+{\left({}^{6}C_2\right)}^2+{\left({}^{6}C_3\right)}^2+{\left({}^{6}C_2\right)}^2+{\left({}^{6}C_1\right)}^2+{\left({}^{6}C_0\right)}^2 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[{\left({}^{6}C_0\right)}^2+{\left({}^{6}C_1\right)}^2+{\left({}^{6}C_2\right)}^2\right]+{\left({}^{6}C_3\right)}^2\left[\because {}^{n}C_r={}^{n}C_{n-r}\right] \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[{\left(\frac{6!}{\left(6-0\right)!\times 0!}\right)}^2+{\left(\frac{6!}{\left(6-1\right)!\times 1!}\right)}^2+{\left(\frac{6!}{\left(6-2\right)!\times 2!}\right)}^2\right]+{\left(\frac{6!}{\left(6-3\right)!\times 3!}\right)}^2 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[{\left(\frac{6!}{\left(6\right)!\times 0!}\right)}^2+{\left(\frac{6!}{\left(5\right)!\times 1!}\right)}^2+{\left(\frac{6!}{\left(4\right)!\times 2!}\right)}^2\right]+{\left(\frac{6!}{\left(3\right)!\times 3!}\right)}^2 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[{\left(\frac{1}{0!}\right)}^2+{\left(\frac{6}{1!}\right)}^2+{\left(\frac{6\times 5}{2!}\right)}^2\right]+{\left(\frac{6\times 5\times 4}{3!}\right)}^2 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[{\left(\frac{1}{1}\right)}^2+{\left(\frac{6}{1}\right)}^2+{\left(\frac{30}{2\times 1}\right)}^2\right]+{\left(\frac{120}{3\times 2\times 1}\right)}^2 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[{\left(1\right)}^2+{\left(6\right)}^2+{\left(15\right)}^2\right]+{\left(20\right)}^2 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=2\left[1+36+225\right]+400 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=\left(2\times 262\right)+400 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=524+400 \\ \Rightarrow \sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)=924 \end{gathered}$$

Therefore, the value of $\sum _{r=0}^6\left({}^{6}C_r·{}^{6}C_{6-r}\right)$ is equal to $924$.

Hence, the correct option is (B).