If ${4.}^{n} C_{6} = {33.}^{n - 3} C_{3}$ then $n$ is equal to

9

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10

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11

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none of these

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Solution

The correct option is A $11$
$4.^{n} C_{6} = 33. (^{n - 3} C_{3})$
$4. (\frac{n!}{(n - 6)! .6!}) = 33. (\frac{(n - 3)!}{3! . (n - 6)!}$
$4 (n (n - 1) (n - 2)) = 33 (4 \times 5 \times 6)$
$n (n^{2} - 3 n + 2) = 33 (30)$
$n^{3} - 3 n^{2} + 2 n - 990 = 0$
Hence the real root that we get is
$n = 11$ .
Hence $n = 11$ .

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