I think it might help if you go back to basics and work from there.
Sometimes it is better to try to work out a few for yourself, without using formulas.
I learned much of what I know from working out combination and perms by hand gambling on horses, which helped quite a bit but here were some gaps.
For example the number of double from ten horses.
I start off
A horse could win any race ie
x000000000
0x00000000
00x0000000
and so on to
00000000x0
000000000x
That's 10 possibilities, now the second horse could win any of the remaining 9 races
ie for the first example, x000000000, it could be
xx0000000
x0x000000
x00x00000
to
x0000000x
So we have 10x9.
However there is a problem as some of those will be duplicates ie you get xx0000000 when the first horse wins the first race, but you also get xx0000000 when the first horse wins the second race, ie from doing all the possibilities of the second horse when the first horse wins race 2, ie 0x0000000, which also will give a line of xx0000000.Perhaps a simpler way of seeing this is to go to a really simple example, of finding the number of doubles with two horses. the answer is obviously one but my method would initially give starting points of
x0 and
0x
The first one has one double xx, as does the second also xx, however it is obvious they are the same, it might not be so easy to see that is a larger number of races because it is not as immediately obvious.
So I need to take this into account, I know there are two possibilities so I need to divide by 2.
If I was doing trebles it would be 10x9x8, then I need to eliminate the dupulicates from 3 horses, so how many are there? We can have 123, 132, 231, 213, 312, 321, that's 6 in total
so I divide by 6, which is 3! (factorial).
So it is easy to get the formula, ie n!/(r!(n-r)!
Knowing how to do that should be useful in other problems.
Actually I just found this, which basically explains the same thing.
http://betterexplained.com/articles/easy-permutations-and-combinations/
A better explanation than mine I think!