Solve the inequation: 5Cn3<Cn+24, n∈N
5Cn3<Cn+24
i.e. 5×n!3!(n−3)!<(n+2)!4!(n+2−4)!
i.e. 5×n!3!(n−3)!<(n+2)!4!(n−2)!
i.e. 5×n!3!(n−3)!<(n+2)(n+1)n!4×3!(n−2)(n−3)!
i.e. 5<(n+2)(n+1)4(n−2)
i.e. 20n−40<n2+3n+2
i.e. 0<n2−17n+42
i.e. n2−17n+42>0
i.e. (x−14)(x−3)>0
i.e. x−14>0 or x−3>0
i.e. x>14 or x>3
⟹ x>14,n∈N