1.

Find n if, 23C3n = 23C2n+3

Answer»

23C3n = 23C2n+3

If nCx = nCy, then either x = y or x = n - y

∴ 3n = 2n + 3 or 3n = 23 – 2n – 3

∴ n = 3 or n = 4



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