Question

An⁵-An³/Cn⁴ ( aranjamente de n luate câte 5 - aranjamente de n luate cite 3 supra la combinari de n luate cite 4 ) ajutatima va rog dau coroana

Answer 1

[n!/(n-5)!-n!/(n-3)!] / n!/(4! x (n-4)!) si dupa simplificare cu n!= [1/(n-5)!-1/(n-3)!]/1/4!(n-4)!=4!(n-4)! / [(n-4)(n-5)-1]/(n-3)! = 4!(n-3)/ [(n-4)(n-5)-1] =4(n-3)/(n^2-9n+19)
Succes in continuare!