Question

Sa se afle n : $C ^{4} n = \frac{5n (n-3)}{6}$

Answer 1

n>_4, altfel combinarile nu au sens, ''>_''=mai mare sau egal

(n!)/[4!x(n-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)

n(n-1)(n-2)(n-3)=20n(n-3), impartim prin: n(n-3)

(n-1)(n-2)=20

n^2 -3n +2-20=0


n^2 -3n -18=0

(n+3)(n-6)=0

n-6=0

n=6