Question

Rezolvati ecuatia: combinari de n luate cate 4 + combinari de n luate cate 3 = 10 aranjamente de n-1 luate cate 2

Answer 1

n!/4!*(n-4)! +n!/3!(n-3)! =10* (n-1)!/(n-3)!
[(n-4)!(n-3)(n-2)(n-1)n]/2*3*4*(n-4)! + [(n-3)!(n-2)(n-1)n]/2*3*(n-3)! = 10*[(n-3)!(n-2)(n-1)]/(n-3)!
se reduc termenii (n-4)!,(n-3)! si ramane
[(n-3)(n-2)(n-1)n]/24 + [(n-2)(n-1)n]/6 = 10[(n-2)(n-1)] //aducem la acelasi numitor
(n-3)(n-2)(n-1)n + 4*(n-2)(n-1)n = 240[(n-2)(n-1)] //reducem (n-2)(n-1) si ramane..
(n-3)n + 4n =240
n²-3n+4n=240
n²+n-240=0
Δ=1+960=31²
n1=(-1+31)/2=15    n2=(-1-31)/2 nu merge fiindca este negativ