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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!

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First problem: For me, (and perhaps soon for you), the easiest way is to note that there are $\dbinom{24}{6}$ equally likely ways to choose $6$ people from the $24$.

Now we count the "favourables." The number of ways to choose $2$ from the first group, $2$ from the second, and $2$ from the third, is $\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}$. For the probability, divide this by $\dbinom{24}{6}$.

Second problem: This time, we are choosing with replacement. To stay close to your approach, there are $24^6$ sequences of length $6$ made up of not necessarily distinct voters. These are all equally likely.

There is some ambiguity in the statement of the problem. Are the $2$ voters of each type have to be (i) not necessarily distinct or (ii) distinct?

(i) Not necessarily distinct: We ask how many strings there are that contain $2$ voters of each type. The positions of the voters of the first type can be chosen in $\dbinom{6}{2}$ ways. For each choice, there are $10^2$ ways to fill these positions with such voters. For each such choice, there are $\dbinom{4}{2}8^2$ ways to deal with the second type of voter. And then there are $\dbinom{2}{2}6^2$ ways for the third type, where of course the $\dbinom{2}{2}$ is superfluous. That gives a total of $\dbinom{6}{2}10^2\dbinom{4}{2}8^2\dbinom{2}{2}6^2$. Divide.

(ii) Distinct: The calculation is very similar, except that $10^2$ is replaced by $(10)(9)$, and $8^2$ by $(8)(7)$, and $6^2$ by $(6)(5)$.

Remarks: Here is an approach to the first problem that is closer in spirit to yours. We can, as you did, observe that there are $(24)(23)(22)(21)(20)(19)$ strings of length $6$ made up of distinct voters. These are all equally likely.

Now we need to count the strings made up of $2$ voters of each type. The particular voters can be chosen in $\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}$ ways. For each such way, there are $6!$ ways to arrange them in a row, giving a total of $\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}6!$ ways. For the probability, divide by $(24)(23)(22)(21)(20)(19)$.

For the second problem, I somewhat prefer to work directly with probabilities. We solve the problem for interpretation (i), the not necessarily distinct case. The analysis for Case (ii) is similar.

We first find the probability that the first $2$ chosen people are of the first type, the next $2$ are of the second type,and the last $2$ of the third type. This probability is $a=\left(\dfrac{10}{24}\right)^2\left(\dfrac{8}{24}\right)^2\left(\dfrac{6}{24}\right)^2$.

But there are $k=\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}$ ways to choose the positions. So our probability is $ka$.

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Ok, when you say divide, you mean by C(24,6), correct? so C(10,2)*C(8,2)*C(6,2)/C(24,6) would be your answer to the first, etc.? – BadAtGraphs Nov 16 '12 at 16:46
Yes, divide by $\binom{24}{6}$. – André Nicolas Nov 16 '12 at 16:51

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