If 2nC3:nC2=44:3, find n.
2n!3!(2n−3)!n!2!(n−2)!=443⇒2n!2!(n−2)!3!(2n−3)!n!=443⇒2n!2!(n−2)3×2×(2n−3)!×n(n−1)(n−2)!=443⇒2n!3n(n−1)(2n−3)!=443⇒2n!=44(n)(n−1)(2n−3)!⇒2n(2n−1)(2n−2)=44n(n−1)⇒(2n−1)(n−1)=11(n−1)Q n=6∴n=6