You good, RainMan?
There are 100*99*98*...*91 = 100! / 90! = 100! / (100-10)! ways to draw 10 balls from 100, denoted P(100,10), if we could distinguish the balls apart. That's the total number of elementary outcomes. Given that we can't distinguish among the white balls or among the red balls, then we can group all these elementary outcomes into 2 ** 10 distinct outcomes where the order would matter.
Here for example, is one of those
WRWRR RRRRW
The above permutation of repeated objects has 3 white orbs and 7 red orbs. Given that I could get 3 white orbs in any of P(50,3) ways from the urn of 50 white balls and I could get 7 red orbs in any of P(50,7) ways, there are
P(50, 3) * P(50, 7 ) ways that I could generate that particular permutation from the elementary outcomes.
Now suppose that I want to group all of the permutations together according to how many unique white balls it has. In other words, I no longer care about the order in which those white ball appear.
Well, each permutation has the same number of elementary possiblities, P(50, 3) * P(50, 7). Note that I could identify a permutation by the location of the white balls. So, for 3 white balls out of 10, there are P(10,3) / 3! distinguishable locations, which we can denote as C(10,3) combinations. So that means that there are C(10, 3) * P(50, 3) * P(50, 7) elementary outcomes that result in a total of 3 white balls out of 10.
If we want to compute the probability of getting 3 white balls out of 10, you just divide the total number of elementary possibilities that yield 3 white balls,
C(10,3) * P(50,3) * P(50, 7) by the total number of elementary possiblities,
P(100, 10 ).
That's how you get the probability mass function for the hypergeometric. It simplifies very nicely to the formula that is expressed simply using combinations, the formula you'll see in authoritative sources.
I hope you kept that asteroid from hitting us. If you break that rock down a little bit and send it toward the Chiefs front office, I'd appreciate it.
