Why do we divide Permutations to get to Combinations? I'm having a hard time reasoning through the formula for combinations $\frac{n!}{k!\left(n-k\right)!}$. I understand the reason for the permutations formula and I know that for the combinations we divide by $k!$ to adjust for the repeated cases, since order does not matter. But what exactly happens when we divide the set of permutations by $k!$ ? 
I know this may seem like a silly question... I just can't take this for granted, lest I miss a chance to apply it correctly. Can you describe what's happening here step by step, sort like debugging a script?

Let $P(n,k)$ denote the number of permutations on $n$ objects taken $k$ at a time. 
Order in this case is important---change the order and you have a different permutation. For example, 1,2,3 and 3,2,1 are but two of the $3!$ ways to arrange (permute) the elements of the set $\lbrace 1, 2, 3 \rbrace$. Moreover, since there are in general $k!$ ways to permute a set of $k$ objects, the way to obtain a formula for the number of combinations on $n$ objects taken $k$ at a time is to divide $P(n,k) = \frac{n!}{(n-k)!}$ by $k!$. 
This yields 
$$C(n,k) = \frac{P(n,k)}{k!} = \frac{n!}{(n-k)!k!}.$$ 
