Question

Combinat de n luate cate 4 = 5n(n-3) supra 6

Answer 1

n≥4, altfel combinarile nu au sens
n!/[4!×(n-4)! =5n(n-3)/6
n(n-1)(n-2)(n-3)/1×2×3×4=5n(n-3)/6
n(n-1)(n-2)(n-3)/24=5n(n-3)/6
n(n-1)(n-2)(n-3)/4=5n(n-3)             ×4  (inmultim cu 4)
n(n-1)(n-2)(n-3)=20n(n-3)             :  n(n-3) (impartim cu n(n-3)
(n-1)(n-2)=20
n² -3n +2=20
n² -3n -18=0
(n+3)(n-6)=0
n-6=0
n=6