Question:
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?
Work:
(A)
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
(B)
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.
Best Answer
Given that there are so many black blocks, there are not many ways to select their positions so that no two are touching. One way to count is to line up eleven blocks with the blacks in the odd number positions. Then you can insert the odd block in $7$ ways, as right of the leftmost block and left of the rightmost block are the same. We can permute the six other blocks in $\frac {6!}{4!1!1!}=30$ ways. So the total number of ways is $7 \cdot 30 = 210$. The probability that one of these is selected is $\frac {210}{27720}$