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Thread: (Basic) Combinatorial Identity.

  1. #1
    Newbie Elvis's Avatar
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    (Basic) Combinatorial Identity.

    {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).
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Elvis View Post
    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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