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If $^{8} C_{3} +^{(n+2)} C_{4} =^{9} C_{4}$ , then n is

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Detailed Solution

$^{8} c_{3} +^{(n + 2)} c_{4} =^{9} c_{4} \Rightarrow n + 2 = 8$ $[\sin c e C (n, r - 1) + C (n, r) = C (n + 1, r)]$

$⇒ n = 6$

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