Suppose that $n$ balls are distributed randomly into $m$ urns. Determine the probability that exactly $k$ balls are in the first urn.
Progress: I treated the balls as indistinguishable and the urns as distinguishable. So this means there are $n+m-1 \choose n$ ways to distribute the balls. I don't know how to find how many ways to get exactly $k$ balls in first urn.
On
We assume that the balls are "thrown" one at a time, and any ball is equally likely to fall in any of the urns, with the results of the throws independent.
Then the probability a thrown ball lands in the first urn is $1/m$. Call landing in the first urn a success. We want the probability of having exactly $k$ successes. The number of successes has binomial distribution, so the probability is $\binom{n}{k}(1/m)^k(1-1/m)^{n-k}$.
First, consider the total number of possibilities, which can be computed through a simple bijection, aka Stars and Bars. For example, if you have to distribute 5 balls into 2 urns, then you could do the following: Let _ correspond to a partition. All *'s before it correspond to the number of balls in the first urn, and the *'s after it correspond to the number of balls in the second. *_**** would correspond to 1 balls first urn 4 balls second urn. You're just counting how many ways the *'s and _'s can be arranged, which is (n+m-1)!/((n!)(m-1)!).
Now, you need to count the number of ways that the remaining balls can be arranged after assuming the first urn has k, since this would give you the favorable outcomes component. This can be calculated thusly: (n-k+m-2)!/((n-k)!(m-2)!) by generalizing the stars and bars method previously illustrated. So now, the answer (unsimplified) is ((n-k+m-2)!/((n-k)!(m-2)!))/ ((n+m-1)!/((n!)(m-1)!))