Question

If ${}^{(a^2-a)}C_2={}^{(a^2-a)}C_4$ , then a =
(a) 2
(b) 3
(c) 4
(d) none of these

Answer 1

(b) 3

$a^2-a=2+4$ [∵${}^n{C}_x{=}^n{C}_y\Rightarrow n=x+y\mathrm{or}x=y$]
$$\begin{gathered} \Rightarrow a^2-a-6=0 \\ \Rightarrow a^2-3a+2a-6=0 \\ \Rightarrow a\left(a-3\right)+2\left(a-3\right)=0 \\ \Rightarrow \left(a+2\right)\left(a-3\right)=0 \end{gathered}$$
$\Rightarrow a=-2\mathrm{or}a=3$
But, $a=-2$ is not possible.
∴ $a=3$