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Question
n का मान निकालिए, यदि `""^(2n)C_2 : ""^nC_2 = 12 : 1`
Sum
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Solution
`(""^(2n)C_3)/(""^nC_3) = 12/1`
⇒ `((2n)!)/(3!(2n - 3)!) xx (3!(n - 3)!)/(n!) = 12/1`
⇒ `((2n)(2n - 1)(2n - 2)(2n - 3)!)/((2n - 3)!) xx ((n - 3)!)/(n(n - 1)(n - 2)(n - 3)!) = 12`
⇒ `(2(2n - 1)(2n - 2))/((n - 1)(n - 2)) = 12`
⇒ `(4(2n - 1)(n - 1))/((n - 1)(n - 2)) = 12`
⇒ `((2n - 1))/((n - 2)) = 3`
⇒ 2n - 1 = 3 (n - 2)
⇒ 2n - 1 = 3n - 6
⇒ 3n - 2n = -1 + 6
⇒ n = 5
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