The problem in full is:
If 10 voters are for A, 8 voters are for B, and 6 voters are for C, what is the probability that a random selection of 6 (no two can be the same) voters will yield 2 voters for each candidate?
and a follow up is
Instead, suppose you call 6 numbers chosen randomly (same number can be chosen) from the pool of 24 voters--now what is the probability that the calling will yield 2 voters for each candidate?
my work so far is as follows:
for the first:
(C(10,2)*C(8,2)*C(6,2) / 3!) / (24*23*22*21*20*19)
and for the second:
(C(10,2)*C(8,2)*C(6,2) / 3!) / (24^6)
my thought process is that you must get 2 from each group, you must divide by 3! to eliminate combinations including the same people in different order, and the denominator is the total number of ways to choose/call 6 people in each different case.
Thanks for any help in advance!