Question

Evaluate ${}^2{C}_2$

Answer 1

We know that

${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

Given ${}^2{C}_2$

Where $n=2,r=2$

$$\begin{gathered} {\therefore }^n{C}_r=\frac{n!}{r!\left(n-r\right)!} \\ =\frac{2!}{2!\left(2-2\right)!} \\ =\frac{2!}{2!\left(0\right)!} \\ but0!=1 \\ =\frac{2!}{2!\times 1} \\ =\frac{2\times 1}{2\times 1\times 1} \\ =\frac{1}{1} \\ =1 \end{gathered}$$

Hence, ${}^2{C}_2$ is $1$