Esame di Stato 2001 – 6

Dimostrare che si ha $\displaystyle {n \choose k} ={n-1 \choose k}+{n-1 \choose k-1}$ dove n, k sono numeri naturali qualsiasi, con n > k > 0.

Osserva

$n!=n(n-1)!$

$\displaystyle {n \choose k} = \frac{n!}{k!(n-k)!}$

Quindi

$\displaystyle {n-1 \choose k}+{n-1 \choose k-1} = \frac{(n-1)!}{k!(n-1-k)!} + \frac{(n-1)!}{(k-1)!(n-1-k+1)!}$

$\displaystyle = \frac{(n-1)!}{k!(n-k-1)!} + \frac{(n-1)!}{(k-1)!(n-k)!}$

$\displaystyle = \frac{(n-1)!}{k(k-1)!(n-k-1)!} + \frac{(n-1)!}{(k-1)!(n-k)(n-k-1)!}$

$\displaystyle = \frac{(n-1)!}{(k-1)!(n-k-1)!} \left(\frac{1}{k}+\frac{1}{n-k}\right)$

$\displaystyle = \frac{(n-1)!}{(k-1)!(n-k-1)!} \left(\frac{n-k+k}{k(n-k)}\right)$

$\displaystyle = \frac{n(n-1)!}{k(k-1)!(n-k)(n-k-1)!}$

$\displaystyle = \frac{n!}{k!(n-k)!}$

$\displaystyle ={n \choose k}$