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Algebra - Elements of Combinatorics

Solve the inequation: 5Cn3<Cn+24, nN


5Cn3<Cn+24

i.e. 5×n!3!(n3)!<(n+2)!4!(n+24)!

i.e. 5×n!3!(n3)!<(n+2)!4!(n2)!

i.e. 5×n!3!(n3)!<(n+2)(n+1)n!4×3!(n2)(n3)!

i.e. 5<(n+2)(n+1)4(n2)

i.e. 20n40<n2+3n+2

i.e. 0<n217n+42

i.e. n217n+42>0

i.e. (x14)(x3)>0

i.e. x14>0 or x3>0

i.e. x>14 or x>3

x>14,nN