if we have 7 black, 5 red and 4 yellow nonstop. what is the probability to pick up 2 black and 1 red without putting them back.
that's 3/16, right? but then I thought that if it's specified wether it's black 1 and black 2, I then found that there would be 6 scenarios. I thought about it and realized that by that logic, every single nonstop would have to be specified, there is 16 nonstop so the total amount of scenarios (not just those who fit under the problem) would be 16! which is 20 922 789 888 000. how would I calculate how many of these scenarios fit under the problem?
+118724 First, what is a nonstop?
for the first scenario you can have BBR, BRB, RBB
$$\mbox{P(2 blacks and a red)=}3\times \frac{7}{16}\times \frac{6}{15} \times \frac{5}{14}=\frac{3}{16}$$
Now I am not sure what the second part of your question is.
Do you want B1, B2, and R1 where all the reds and all the blacks have different numbers?
In this case it would be
$$\mbox{P(B1, B2 and R1)= }6\times \frac{1}{16}\times \frac{1}{15} \times \frac{1}{14}=\frac{6}{3360}=\frac{1}{560}$$
I am not particularly good at probability but I think that is correct.
-----------------------------------------------------------
16! is the number of different ways the 16 nonstops can be ordered. Assuming that they are all different from each other. We don't care about that, we are only concerned with 3 of them.
+118724 I just want to think about this some more.
The number of combinations of 3 from 16 is 16C3 = 560
We want 1 particular combination so the probability will be $${\frac{{\mathtt{1}}}{{\mathtt{560}}}}$$
Different method of attack. Same answer.
I think I'm onto something here! 
