InterviewSolution
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InterviewSolution
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यदि `((2n)!)/(3!(2n-3)!)` तथा `(n!)/(2!(n-2)!), 44:3` के अनुपात में हैं, तो n का मान ज्ञात कीजिए । |
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Answer» प्रश्नानुसार `((((2n)!)/(3!(2n-3)!)))/(((n!)/(2!(n-2)!)) ) = 44/3` `rArr ((2n)!2!(n-2)!)/(3!(2n-3)!n!) = 44/3` `rArr (2n(2n-1)(2n-2)(2n-3)!)/(3!(2n-3)!) .(2!(n-2)!)/(n(n-1)(n-2)!)= 44/3` `rArr(2n(2n-1)(2n-2))/(1.2.3) xx (1.2)/(n(n-1)) = 44/3` `rArr (2n-1)(2n-2) = 22(n-1)` `rArr 4n^(2) - 6n+ 2 = 22n - 22` `rArr 4n^(2) -28+24 = 0` `rArr n^(2 ) - 7n+ 6 =0 rArr(n-1) (n-6) = 0` `rArr n = 1` या `n = 6` |
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