InterviewSolution
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If `.^(n)C_(8) = .^(n)C_(2)`, find `.^(n)C_(2)`. |
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Answer» `.^(n)C_(x) = .^(n)C_(Y)` `rArr n = x +y` `:. .^(n)C_(8) = .^(n)C_(2)` `rArr n = 8 +2 = 10` Therefore, `.^(n)C_(2) = .^(10)C_(2) = (10!)/(2!8!) = (10 xx 9 xx 8!)/(2xx 1 xx 8!) = 45`. |
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