Why do combinations have each element appear an equal number of times? Say you have a set s = {1, 2, 3}. All of the possible combinations of that set would be 12, 13, 23. Note that there are 2 1s, 2 2s, and 2 3s. Is there any way to prove that this is true for any set and for combinations of any length?
 A: No element is favoured over any other, so by symmetry each must appear equally often.
A: Suppose you have a set with N elements {1,...,N}. The number of two-element combinations including 1 is N-1 -- we have {12, 13,..., 1N}. The number of two element combinations including 2 is N-1 -- we have N-2 in {23, 24,..., 2N} plus one from the 1s (12). Lather, rinse, repeat for all the two-elements. Then the same idea holds for the three-elements, four-elements, ... , and N-elements.  
Edit to add higher case logic:
Yes, you would have 123, 124, ..., 12N, 134, ..., 1(N-1)N, ... (N-2)(N-1)N. So, the question then becomes how many of each number are there? Let's go a bit slower this time though, and instead of asking how many include 1, let's ask how many go 12_. Well it's N-2; 123, 124, ..., 12N. Then there are N-3 that go 13_, and.... Same idea for the rest of the 1__s, giving a total of (N-2)+(N-3)...+1 combinations with a 1.  
We could keep going at this point, but there's an easier way to think about it. Get a piece of paper and write out the combinations in triangles, like this:
123 124 125 126 127 128 129
.......134 135 136 137 138 139
..............145 146 147 148 149
         ....
........................................189
.......234 235 236 237 238 239
..............235 236 237 238 239
....
.........................................289
You get the  idea. Increasing last numbers to the right, increasing middle numbers down, grouped by first number. Formatting is hard though, and I'm lazy, so I'll leave writing the rest as an exercise for the reader :-)
Specifically, notice the shapes and what's happening to them. They form triangles that consistently get smaller - the triangle of combinations starting with 2 is one smaller both vertically and horizontally than the triangle of combinations starting with 1. The triangle starting with 1 has N-2 as its top (remember the 12_s that we calculated?), then N-3 in its second row (the 13_s), all the way down to 1(N-1)N. Meanwhile, the triangle starting with 2 has N-3 at its top, then N-4, all the way down to a single 2(N-1)N.
But wait a second! There were N-2 combinations with a 2 in the top triangle -- the 12_s. So when we're counting our total number of combinations with a 2, we have to count those too. That makes our total of combinations with a 2 to be (N-2)+(N-3)+...+1. Now doesn't that look familiar? It's the same as the number of combinations with a 1!
We quickly see that it's the same for all of the numbers. The combinations with 3s get N-2 from the 1-triangle, N-3 from the 2-triangle, and the remaining (N-4)+...+1 from the 3-triangle. Going all the way down, the combinations with N get 1 from the N-triangle, 2 from the (N-2)-triangle, 3 from the.... Again, you get the idea.
This way of thinking about it should be much easier to extrapolate into larger groupings.
