(Basic) Combinatorial Identity.

**Elvis** · February 1st 2010, 09:59 PM

${n\choose{k}}+{n\choose{k-1}} = \frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}$ = ${\color{blue}\frac{n!}{k!(n-k+1)!}\left\{(n-k+1)+k\right\}} = {{n+1}\choose{k}}$ .

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

**TheEmptySet** · February 1st 2010, 10:12 PM

Originally Posted by Elvis

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

get a common denominator

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

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