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Question

If (2n)!3!(2n3)!andn!2!(n2)! are in the ratio 44:3, find n.

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Solution

We have,
(2n)!3!(2n3)!n!2!(n2)!=443(2n)!×2!(n2)!3!(2n3)!×n!=443(2n)(2n1)(2n2)(2n3)!×2!(n2)!3×2!(2n3)!×n(n1)(n2)!=4432n(2n1)(2n2)3n(n1)=4432(2n1)×2(n1)3(n1)=4434(2n1)=442n1=112n=12n=6n=6


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