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24. Determine n if (i) 2nC3 : nC3 = 12 : 1 (ii) 2nC3 : nC3 = 11 : 1

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    24. i. 2nC3 : nC3 = 12 : 1

    => 2n!3!(2n−3)! ÷ n!3!(n−3)! = 121

    => 2n(2n−1)(2n−2)(2n−3)!3!(2n−3)! × 3!(n−3)!n(n−1)(n−2)(n−3)! = 12

    => 2n(2n−1)(2n−2)n(n−1)(n−2) = 12

    => 2(2n−1)2(n−1)(n−1)(n−2) = 12

    => 4(2n - 1) = 12(n – 2)

    => 8n – 4 = 12n – 24

    => 24 – 4 = 12n – 8n

    => 20 = 4n

    =>n = 204

    =>n = 5

    ii. 2nC3 : nC3 = 11 : 1

    => 2n!3!(2n−3)! ÷ n!3!(n−3)! = 111

    => 2n(2n−1)(2n−2)(2n−3)!3!(2n−3)! × 3!(n−3)!n(n−1)(n−2)(n−3)! = 11

    => 2n(2n−1)2(n−1)n(n−1)(n−2) = 11

    => 4(2n – 1) = 11(n – 2)

    => 8n – 4 = 11n – 22

    => 22 – 4 = 11n – 8n

    => 18 = 3n

    =>n = 183

    =>n = 6

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