Question:
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
Solution:
(b) 3
$a^{2}-a=2+4 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$
$\Rightarrow a^{2}-a-6=0$
$\Rightarrow a^{2}-3 a+2 a-6=0$
$\Rightarrow a(a-3)+2(a-3)=0$
$\Rightarrow(a+2)(a-3)=0$
$\Rightarrow a=-2$ or $a=3$
But, $a=-2$ is not possible.
$\therefore a=3$
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