"Need a few good math sites to explain how to work out the following: The number of combinations (not really combinations) that you could set up pool balls in, 7 red, 7 yellow, and 1 black."
Yes, you obviously mean
arrangements.
We have
n kinds of objects (e.g. 3 categories of balls), for each of which there are an equal
m number of copies (like 7 reds, 7 yellows, 7 blacks, etc). Then the number of all possible arrangements of those objects is
(n*m)! / (m!)^n
To answer your question, leave the one black ball out for a moment. Plugging in n=2 kinds of balls, and m=7 of each kind, i.e. 7 reds and 7 yellows, we get 3432 possible arrangements of the 14 red & yellow balls.
There's still the one black ball left. The black ball can be placed anywhere in an arrangement of reds & yellows. Each such arrangement contains 14 balls in all, so the black ball can be placed in any of the 15 available positions (there's one extra position because the black ball can go at the front as well, i.e. in position zero). So 15*3432=
51,480 is the answer.
For more complicated calculations, e.g. unlimited repetition of objects of n kinds and no restriction on the number of times any object can appear, one needs enumerating generating functions. Try
www.mathworld.com
I don't know of websites when you can plug in your numbers for such problems and get the answer. Sorry.