If $^{k^{2} - K} C_{2} =^{k^{2} - k} C_{4}$ , then k=

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none of these

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Solution

The correct option is A 3
We know that if $^{n} C_{p} =^{n} C_{q}$ then p + q = n
$∴ 2 + 4 = k^{2} - k$ or $k^{2} - k - 6 = 0$ or (k - 3) (k + 2) =0
$∴$ k =3, -2 but -2 is -ve so rejected.
$∴ k = 3 ⟹ (b)$

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