My apologies, thanks for the corrections!
The biggest problem I'm having has to do with, for example, the need for two green balls or two red balls in each sample in the first question, or the need for three diamonds or black cards. I know that the number of combinations is
C(n,r) = n!/(r!(n-r)!)
where n is the number of distinct objects and r is the number taken at a time.
For the first problem, I know that if you take out two green balls, there would be 6 left, but since you can only have two in one hand, does that mean that you really only have 9 balls left to use as your n? Likewise, for the red balls, you would have 14 left. I then figured that you would multiply the two combinations together, giving 252,252, but that seems unreasonably high. It certainly is too high.
How many ways in total can you select 5 balls from 17. The answer is
17! / [(17 - 5)! * 5!] = 17 * 16 * 15 * 14 * 13 / (5 * 4 * 3 * 2) =
17 * 16 * 15 * 14 * 13 / (15 * 8) = 17 * 2 * 1 * 14 * 13 = 6,188, way less than 252,252
Imagine the balls have numbers on them. So green balls are numbered 1 through 8, pink balls are numbered 1 through 6, and red balls are numbered 1 through 3.
How many ways can you select two green balls from eight. You do not care about order; picking green 3 first and green 5 second is the same to you as picking green 5 first and green 3 second. So the answer is 8! / [(8 - 2)! * 2!] = 8! / (6! * 2!) = 7 * 8 / 2 = 28.
How many ways can you select two pink balls from six. By the same logic, 6! /[(6 - 2!) * 2!] = 6! / (4! * 2!) = 6 * 5 / 2 =15.
How many ways can you select one red ball from 3? Obviously 3! / [(3 - 1)! * 1!] = 3! / (2! * 1!) = 3 / 1 = 3.
So, for each possibility of pink balls, you have three possibilites for red balls, giving 3 * 15 = 45 possibilities for pink and red balls.
And for each possibility of red and pink balls, you have 28 possibilities for green balls, giving 45 * 28 = 1260.
I'm curious about that reasoning for the other two problems as well.
Does this seem right, or are my suspicions right?