There are $10$ marbles, $4$ blue and $6$ red There are $10$ marbles, $4$ blue and $6$ red. $6$ are chosen. What's the probability that exactly $5$ are red?
My problem is I don't know if order matters or doesn't:
I think first I have to chose $5$ red out of $6$: ${6 \choose 5}$. I don't think the order in which we chose the $5$ matters because if I chose $R_1,R_2,...R_5$ it's the same thing if I chose $R_5,...R_1$. Or does the order matter? 
Next  I need to chose $1$ blue out of $4$. So I think the answer is,
$$\frac{{6 \choose 5} {4 \choose 1}}{{10 \choose 6}}$$
 A: The order does not matter, the subset of red marbles are indistinguishable, as is the subset of blue. 
This is an example of sampling from a hypergeometric distribution.
A: To understand that order of picking being treated as important doesn't 
change the answer, consider that the balls were numbered (distinguishable) and you later had to arrange the balls (say in a line).This is the same
as counting the same choice of 
balls as different if the order in which they were picked were different.
But then, the point is, the total number of ways to pick in that case 
(your denominator) would also have to change in order to reflect the fact
that the order of picking is important. Again, just think of picking the 
balls and then permuting them.
So, in this case, the required probability would be:
$$ \dfrac{{6 \choose 5}{4 \choose 1} \times 6!}{{10 \choose 6} \times 6!}$$.
So, the answer is the same as before, if you introduced the condition that
the order of picking is important.
A: Yes, you've got it right. The order doesn't matter.
