Alice, Bob, and Carol sample uniformly at random from a bucket which has 10 blue balls, and 6 red balls, until the bucket is empty. What is the probability that Alice picks 3 blue balls, Bob picks 5 blue balls, and Carol picks 2 blue balls?
I know how to answer this one way, using the multinomial distribution: There are $\frac{10!}{3!5!2!}$ such arrangements, and each ball has probability $\frac{1}{3}$ of being picked by any given person, so the answer is $\frac{10!}{3!5!2!}\cdot\frac{1}{3^{10}}$
Why doesn't the following approach work instead? : Treat this as a star and bars problem, for 10 identical balls and 3 distinguishable bins. Then, there are (10+2)C2 = 12C2 = 66 possible arrangements of balls in bins. Of these, there is only one such that bin 1 has 3 balls, bin 2 has 5 balls, and bin 3 has 2 balls. But the answer is obviously not $\frac{1}{66}$.