Question

Determinați n€N, n≥2, pt care combinari de n luate cate 2 = 4 aranjamentede n luate cate 1

Answer 1

Conditii de existenta n≥2
n(n-1)/2=4n
n²-n=8n
n²-9n=0
n(n-9)=0
n=0 nu are sens
n=9

berificare 9*8/2=4*9
36=36 adevarat, bine rezolvat

Answer 2

C de n luate cate 2=4(n!/(n-1)!)

n!/[(2!(n-2)!]=4n
[(n-1)n]/2=4n
(n-1)n=8n
n^2-n=8n
n^2-9n=0
n(n-9)=0
1. n=0 FALS PTC n> sau egal 2
2. n-9=0
n=9 ADEVARAT