If $^{35} C_{n + 7} =^{35} C_{4 n - 2}$ , then write the values of n.

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Solution

$^{35} C_{n + 7} =^{35} C_{4 n - 2}$

$\Rightarrow n + 7 = 4 n - 2$ $\Rightarrow 3 n = 9$ $\Rightarrow n = 3$ Or $^{35} C_{(n + 7)} =^{35} C_{(4 n - 2)}$

$\Rightarrow 35 - (n + 7) = 4 x - 2$ $\Rightarrow 35 - x - 7 = 4 x - 2$

$\Rightarrow 5 x = 30$ $\Rightarrow n = 6$

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