Question

If $n=5$, then ${\left({}^{n}c_0\right)}^2+{\left({}^{n}c_1\right)}^2+{\left({}^{n}c_2\right)}^2+..+{\left({}^{n}c_5\right)}^2$

• A. $250$
• B. $254$
• C. $245$
• D. $252$
• E. $258$

Answer 1

The correct option is D

$252$

Explanation for correct option

Given $n=5$

$$\begin{gathered} \therefore {\left({}^{n}c_0\right)}^2+{\left({}^{n}c_1\right)}^2+{\left({}^{n}c_2\right)}^2+..+{\left({}^{n}c_5\right)}^2 \\ ={\left({}^{5}c_0\right)}^2+{\left({}^{5}c_1\right)}^2+{\left({}^{5}c_2\right)}^2+{\left({}^{5}c_3\right)}^2+{\left({}^{5}c_4\right)}^2+{\left({}^{5}c_5\right)}^2\left[\because n=5\right] \\ ={\left(\frac{5!}{0!\left(5-0\right)!}\right)}^2+{\left(\frac{5!}{1!\left(5-1\right)!}\right)}^2+{\left(\frac{5!}{2!\left(5-2\right)!}\right)}^2+{\left(\frac{5!}{3!\left(5-3\right)!}\right)}^2+{\left(\frac{5!}{4!\left(5-4\right)!}\right)}^2+{\left(\frac{5!}{5!\left(5-5\right)!}\right)}^2\left[\because {}^{n}c_m=\frac{n!}{m!\left(n-m\right)!}\right] \\ ={\left(1\right)}^2+{\left(5\right)}^2+{\left(10\right)}^2+{\left(10\right)}^2+{\left(5\right)}^2+\left(1\right) \\ =1+25+100+100+25+1 \\ =252 \end{gathered}$$

Hence, option(D) i.e. $252$ is correct