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Question

If 2nC3:nC2=44:3, find n.

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Solution

2n!3!(2n3)!n!2!(n2)!=4432n!2!(n2)!3!(2n3)!n!=4432n!2!(n2)3×2×(2n3)!×n(n1)(n2)!=4432n!3n(n1)(2n3)!=4432n!=44(n)(n1)(2n3)!2n(2n1)(2n2)=44n(n1)(2n1)(n1)=11(n1)Q n=6n=6


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