Solve the following


If ${ }^{\left(a^{2}-a\right)} C_{2}={ }^{\left(a^{2}-a\right)} C_{4}$, then $a=$

(a) 2

(b) 3

(c) 4

(d) none of these


(b) 3

$a^{2}-a=2+4 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$

$\Rightarrow a^{2}-a-6=0$


$\Rightarrow a^{2}-3 a+2 a-6=0$

$\Rightarrow a(a-3)+2(a-3)=0$



$\Rightarrow a=-2$ or $a=3$

But, $a=-2$ is not possible.

$\therefore a=3$

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