If $^{n} C_{3} =^{n} C_{5^{'}}$ then find the value of n:

9

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10

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8

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7

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Solution

The correct option is C $8$
$^{n} C_{r} = \frac{n!}{(r!) (n - r)!}$
Given, $^{n} C_{3} =^{n} C_{5}$
$=> \frac{n!}{(3!) (n - 3)!} = \frac{n!}{5! (n - 5)!}$
$=> 3! \times (n - 3) \times (n - 4) \times (n - 5)! = 5 \times 4 \times 3! \times (n - 5)!$
$=> (n - 3) \times (n - 4) = 5 \times 4$
From this, $n - 3 = 5$
$=> n = 8$

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