Se impune conditia x∈N, x-2>2; x>4
[tex]C_{x-2}^2 =21[/tex]
[tex] \frac{(x-2)!}{2(x-4)!}=21 [/tex]
[tex] \frac{(x-2)(x-3)(x-4)!}{2(x-4)!} =21[/tex]
(x-2)(x-3)=42
x²-5x+6=42
x²-5x-38=0
Δ=(-5)²-(-38*4)=169
[tex] x_{1}= \frac{5+13}{2}=9 [/tex]
De a doua solutie a ecuatiei nu avem nevoie deoarece e numar negativ si nu verifica conditia impusa la inceput.
Deci x=9
