# (Basic) Combinatorial Identity.

• Feb 1st 2010, 09:59 PM
Elvis
(Basic) Combinatorial Identity.
Quote:

$\displaystyle {n\choose{k}}+{n\choose{k-1}} = \frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}$ = $\displaystyle {\color{blue}\frac{n!}{k!(n-k+1)!}\left\{(n-k+1)+k\right\}} = {{n+1}\choose{k}}$.
Can someone explain the blue step, please? I can't follow how $\displaystyle \frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}$ is manipulated to give $\displaystyle \frac{n!}{k!(n-k+1)!}\left\{(n-k+1)+k\right\}$. (I understand the last step).
• Feb 1st 2010, 10:12 PM
TheEmptySet
Quote:

Originally Posted by Elvis
Can someone explain the blue step, please? I can't follow how $\displaystyle \frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}$ is manipulated to give $\displaystyle \frac{n!}{k!(n-k+1)!}\left\{(n-k+1)+k\right\}$. (I understand the last step).

get a common denominator

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