<s>[INST] 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. 
 [/INST] We have the identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$. So we see the series telescope:
$$\sum_{i=0}^{k} (-1)^{i} \binom{n}{i} = \binom{n}{0} + \sum_{i=1}^{k} \binom{n-1}{i-1} + \sum_{i=0}^{k} \binom{n-1}{i}$$ 
So we see $\binom{n}{0} = 1$. Then $\binom{n}{1} = \binom{n-1}{0} + \binom{n-1}{1}$. For any $x$, $\binom{x}{0} = 1$. So $\binom{n}{0} - \binom{n-1}{0} = 0$.
Now look at $\binom{n}{2} = \binom{n-1}{1} + \binom{n-2}{2}$. By telescoping, $-\binom{n-1}{1} + \binom{n-1}{1} = 0$.
So we are left with the term $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$, with the $\binom{n-1}{k-1}$ cancelling out out the $\binom{n-1}{k-1}$ term from $\binom{n}{k-1}$. Then we are left with $\binom{n-1}{k}$. 
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