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NCERT Class XI Mathematics - Permutations and Combinations - Solutions

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Question : 23 of 41
Marks: +1, -0
Determine n if
(i)  2nC3\,{}^{2n}C_3 :  nc3\,{}^{n}c_3 = 12 : 1 (ii)  2nC3\,{}^{2n}C_3 :  nC3\,{}^{n}C_3 = 11 : 11
Solution:  
(i)  2nC3\,{}^{2n}C_3 :  nc3\,{}^{n}c_3 = 12 : 1
⇒  2nC3 nC3\frac{\,{}^{2n}C_3}{\,{}^{n}C_3} = 121\frac{12}{1} ⇒ 2n!3!(2n−3)!\frac{2n!}{3!(2n-3)!} × 3!(n−3)!n!\frac{3!(n-3)!}{n!} = 12
⇒ 2n(2n−1)(2n−2)n(n−1)(n−2)\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 12 ⇒ 4(2n−1)n−2\frac{4(2n-1)}{n-2} = 12 ⇒ 8n - 4 = 12 n - 24 ⇒ n = 5
(ii)  2nC3\,{}^{2n}C_3 :  nC3\,{}^{n}C_3 = 11 : 11
⇒  2nC3 nC3\frac{\,{}^{2n}C_3}{\,{}^{n}C_3} = 11 ⇒ 2n!3!(2n−3)!\frac{2n!}{3!(2n-3)!} × 3!(n−3)!n!\frac{3!(n-3)!}{n!} = 11
⇒ 2n(2n−1)(2n−2)n(n−1)(n−2)\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 11 ⇒ 4(2n−1)n−2\frac{4(2n-1)}{n-2} = 11 ⇒ 8n - 4 = 11 n - 22 ⇒ n = 6
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