Permutations and Combinations Show that $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ...+(-1)^k * \binom{n}{k} = (-1)^k * \binom{n-1}{k}$.
I know this has to do with permutations and combination problems, but I'm not sure how would I start with this problem. 

I assume that $n \geq 1$ and $K \geq 0$. We need to prove that
$\sum_{k=0}^{K}(-1)^k\binom{n}{k}=(-1)^K\cdot\binom{n-1}{K}$
I like proofs by induction. So we fix some $n \geq 1$ and our base case is $K = 0$. In this case we have (recall that for any $~n~$ $\binom{n}{0} = 1$):
$LHS: ~~ \sum_{k=0}^{0}(-1)^k\binom{n}{k} = (-1)^0\binom{n}{0} = 1\cdot 1=1$
$RHS: ~~ (-1)^0\cdot \binom{n-1}{0} = 1\cdot1=1$
Now we assume that the formula is correct for some $K \geq 0$ and show that then it's correct for $K+1$ (notice that for any $~x~$ and any $~k~$  $~(-1)^{k+1}x=-(-1)^kx~$ thus$~(-1)^kx + (-1)^{k+1}x = 0$):
$\sum_{k=0}^{K+1}(-1)^k\binom{n}{k} = \sum_{k=0}^{K}(-1)^k\binom{n}{k} + (-1)^{K+1}\binom{n}{K+1}=(-1)^K\cdot\binom{n-1}{K}+(-1)^{K+1}\binom{n}{K+1}=(-1)^K\cdot\binom{n-1}{K} + (-1)^{K+1}\binom{n-1}{K} + (-1)^{K+1}\binom{n-1}{K+1} = (-1)^{K+1}\binom{n-1}{K+1}$
