If $^{40} C_{n + 7} =^{40} C_{4 n - 2},$ then all the values of n are given by

28

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3, 6

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3, 7

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6

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Solution

The correct option is C $3, 7$
Here there are two conditions
$n + 7 = 4 n - 2$
$⟹ 3 n = 9$
$⟹ n = 3$
and,
$n + 7 + 4 n - 2 = 40$
$⟹ 5 n + 5 = 40$
$5 n = 35$
$n = 7$
There are two possible value of $n$ . They are $3$ and $7$

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