+130555 ultmage11234's answer is correct....if we consider this to be a combination problem...
If we consider it to be a permute....we have
P(9,3) - P(6,3) = 384 ways
This is a little more difficult to see....
Consider the set of permutes that include one black ball...
If we let it occupy the first position in the set, we have 6 other b***s to choose from in the second position and 5 more to choose from in the 3rd position. Thus, 3 x 6 x 5 = 90
And since a black ball can occupy any one of three positions, this gives 270 ways to permute a set which includes one black ball.
For a set that includes two black b***s, we have 3 ways to choose one for the first position and two ways to choose one for the second position and 6 ways to choose a final ball. But the non-black ball can occupy any one of three positions. So the total number of permutes in a set containing 2 black b***s = 3 x 2 x 6 x 3 = 108
And we have 6 ways to arrange 3 black b***s in a set that just contains 3 black b***s.
So...the total number of permutes of all the sets containing at least one black ball is just
270 + 108 + 6 = 384 .....just as we speculated!!!
{Constructive criticism is welcome.....since I am not a probability "expert" }
