Why does dividing by the number of ways to arrange $n$ things work? Combination formula is defined as:
$$\binom{n}{r}=\frac{n!}{(n-r)!\cdot r!}$$
We do $(n-r)!$ because we want the combinations of only $r$ objects given $n$. We do $r!$ again because in combinations, order does not matter so doing this gets rid of repeated results. But why? Why does $r!$ get rid of anything.
Any visual help is appreciated.
thanks
 A: Because  there are exactly $r!$ arrangements  with the same underlying set of $r$ objects, and we can apply the shepherd's principle.
A: After you select a subset r from n elements, you are left with n-r elements.  It would be considered a different grouping of each r elements for each different way of ordering the n-r elements if order mattered.  Since order doesn't matter for r, we divide by r!, since each combination of r elements will be present in the possibility space r! times - and we only want unique instances of each order.  Then, since the order of the remaining elements doesn't matter either, we divide by (n-r)!.  Because if the ordering of those elements did matter (for example, if we were wondering how many ways we could permute the remaining elements after removing a specific set of r of them), then we would have to multiply the result by (n-r)!
A: The formula $$\displaystyle {n \choose r} = \frac{n!}{(n-r)! \cdot r!},$$ in some sense, tells us how to start with the $n!$ permutations of $\{1, 2, \ldots, n\}$ and ignore certain differences to glue certain sets of permutations together to form ${n \choose r}$ groups (integer division naturally corresponds to gluing things together: $12 \div 4 = 3$ precisely because we can take $12$ things and glue sets of $4$ things together into one group, obtaining $3$ groups).
Maybe the reverse process, `de-grouping,' will be easier to understand; let's rearrange the equation, and think about why ${n \choose r} \cdot (n - r)! \cdot r! = n!$. We'll see how we can start with the ${n \choose r}$ $r$-element subsets and generate all of the $n!$ permutations (and I think you'll be able to see how to do the gluing, after all is said and done).
Let's just pick a concrete example, say, ${5 \choose 2}$. Let's see how each $2$-element subset of $\{1, \ldots, 5\}$ can be `blown up' $2! \cdot (5 - 2)!$ ways, each giving a permutation of $\{1, \ldots, 5\}$.
We'll pick a particular $2$-element subset, say $\{1, 3\}$. Of course, as sets, $\{1, 3\}$ and $\{3, 1\}$ are the same. But we're going from sets to permutations, so we'll get $2!$ (ordered!) lists from the single set $\{1, 3\}$, the list $1,\, 3$ and $3,\, 1$.
We've associated a single $2$-element subset of $\{1, \ldots, 5\}$ to $2!$ lists, but we still don't have a permutation of $\{1, \ldots, 5\}$. Namely, we have yet to incorporate the $n - r = 5 - 2$ elements of $\{2, 4, 5\}$.
To each of our lists $1,\,3$ and $3,\,1$, we append the $(5 - 2)!$ orderings of $\{2, 4, 5\}$ to the end of the list:
\begin{array}{ll|lll c ll|lll}
1 & 3 & 2 & 4 & 5; &\qquad& 3 & 1 & 2 & 4 & 5 \\
1 & 3 & 2 & 5 & 4; &\qquad& 3 & 1 & 2 & 5 & 4 \\
1 & 3 & 4 & 2 & 5; &\qquad& 3 & 1 & 4 & 2 & 5 \\
1 & 3 & 4 & 5 & 2; &\qquad& 3 & 1 & 4 & 5 & 2 \\
1 & 3 & 5 & 2 & 4; &\qquad& 3 & 1 & 5 & 2 & 4 \\
1 & 3 & 5 & 4 & 2; &\qquad& 3 & 1 & 5 & 4 & 2 \\
\end{array}
In this way, we've associated the $2$-element subset $\{1,3\}$ with $2! \cdot (5 - 2)!$ permutations of $\{1,\ldots, 5\}$; all of those permutations whose first two elements are taken from $\{1, 3\}$. Taken in reverse, all of the $2! \cdot (5 - 2)!$ permutations above get glued together to form the $2$-element subset $\{1,3\}$.
If we play this game with each of the ${5 \choose 2}$ $2$-element subsets of $\{1, \ldots, 5\}$, we'll naturally generate ${5 \choose 2} \cdot 2! \cdot (5 - 2)!$ permutations of $\{1, \ldots, 5\}$. 
On the other hand, they'll be all permutations of $\{1, \ldots, 5\}$, since the first two numbers in any permutation of $\{1, \ldots, 5\}$ form a $2$- element subset of $\{1, \ldots, 5\}$. Hence ${5 \choose 2} \cdot (5 - 2)! \cdot 2! = 5!$, and more generally, ${n \choose r} \cdot (n - r)! \cdot r! = n!$.
