# [SOLVED] Proving Combinations

• March 25th 2009, 02:35 PM
DaCoo911
[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)
• March 25th 2009, 02:55 PM
o_O
$C(k, r+1) = \frac{k!}{(r+1)!(k - (r+1))!} = \frac{k!}{(r+1)!(k- r - 1)!}$

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

Can you show that the two are equal?