I've been trying to prove this for quite some time today, I just can't seem to get my head around this one! (Crying) O NO!

C(k, r+1) = C(k, k-r-1)

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- Mar 25th 2009, 02:35 PMDaCoo911[SOLVED] Proving Combinations
I've been trying to prove this for quite some time today, I just can't seem to get my head around this one! (Crying) O NO!

C(k, r+1) = C(k, k-r-1) - Mar 25th 2009, 02:55 PMo_O
$\displaystyle C(k, r+1) = \frac{k!}{(r+1)!(k - (r+1))!} = \frac{k!}{(r+1)!(k- r - 1)!}$

$\displaystyle C(k, k-r-1) = \frac{k!}{(k-r-1)!(k - \left(k-r-1\right))!} = \cdots$

Can you show that the two are equal?