A child has 12 blocks. 6 are black, 4 are red, 1 is white, and 1 is yellow
(A) If the child puts the blocks in a line, how many different arrangements are possible?
(B) If one of the arrangements in (A) is randomly selected, what is the probability that no two black blocks are next to each other?
If we assume the blocks are numbered, 1-12, then there are 12! arrangements possible. Since it isn't mentioned, there are 6! ways to arrange the black, 4! for the red, and 1! for each the white and the yellow.
12!/6!4!!1!1 = 27'720
I'm not sure how to go about (B). I think the probability will be
x/27'720 but I'm not sure how to arrive at x.