Question: Randomly distribute r balls in n boxes. Find the probability that the first box is empty.
I think I should make the question into 3 cases, namely, r=n, r<n, and r>n.
CASE r=n:
${r \choose 2}$ because there are two balls of r balls that will be put into the same box.
${n-1 \choose 1}$ because we choose one box that will put the two balls.
$(r-2)!$ put the rest r-2 balls into the rest n-2 boxes.
$$\frac{{r \choose 2} {n-1 \choose 1} (r-2)! }{r^{n}}$$
CASE r>n: $$(\frac{n-1}{n})^r$$
CASE r<n: $$\frac{{(n-1)Pr}}{nPr}$$
Am I on the right track? Please give me some advice, thank you!