Question

cum pot sa rezolv $C^{n+1}_{n+2} = 2$

Answer 1

(n+2)!/(n+1)!(n+2-n-1)=2
(n+2)!/(n+1)!1!=2
(n+2)(n+1)n/(n+1)n=2
(n+2)/1=2
n+2=2
n=0

Answer 2

Folosim formula combinărilor complementare:

$\it C^k_n =C^{n-k}_n$

$\it C^{n+1}_{n+2} =C^{n+2-n-1}_{n+2} =C^1_{n+2} =n+2$

Ecuatia devine:

n+2 = 2 ⇒ n = 0 .