There are 8 buckets, each bucket is a different color (for simplicity, let's label the colors A, B, C, D, E, F, G and H; if you like: Aqua, Brown, Cyan, Diamond, Eggshell, Fuchsia, Green, Hot-pink). There are 8 balls, each a different colour (also A, B, C, D, E, F, G, H).
Someone randomly throws all 8 balls into all 8 buckets, such that all buckets have exactly 1 ball in them after all 8 balls are thrown, and the process of throwing the balls into each bucket is completely random.
Q: What's the probability of exactly 3 colour matches? That is, what's the probability that 3 (and only 3) buckets have a ball tossed in them that is the same color as the bucket?
Here is what I've done so-far:
The no. of sample points in the sample space is P(8,8) = 8!.
The number of sample points in the event of exactly 3 color matches is C(8,3) (no. of ways of chooisng the 3 'matched' colors of 8), multiplied by the number of ways of arranging the 'unmatched colors' is (8 - 3 - 1 )! = 4! (For example, suppose the unmatched colors are D, E, F, G and H: The color of the ball in the D bucket can be E, F, G or H, i.e. 4 choices to prevent a color match. The number of choices to prevent a match in bucket E will be 3, then 2 in F, then 1 in G.)
So, the answer I have is: C(8,3)*4! / 8!.
One possible issue is with the calculation of 4! To follow from the example in the previous paragraph, suppose an Eggplant ball is placed in the Diamond bucket. Now, there are 4 choices on the second 'unmatched' throw rather than 3... And I don't know how to resolve this. I'm also not sure if the '3 from 8' color matches should be a combination as written, or permutation.
If anyone has any thoughts on this problem -- suggestions for improvement or validation of what I've done -- I would really appreciate it. Incidentally, the problem is based on Example 2.22 from Wackerly et. al.'s 'Mathematical Statistics with Applications' 7ed, 2007, but an extension (they use 3 and 1 instead of 8 and 3), and using a different method from what was done in the example.