An 8-sided die is rolled 15 times. What is the probability that a 2 appears exactly 2 times, a 3 appears exactly 3 times, …, and a 5 appears exactly 5 times?
I tried doing it using binomial probability taking (1/6)^2*(5/6)^13 etc but this seems tedious.
Think of it like this. Each time you roll the dice 15 times you generate a sequence of 15 digits, e.g. (1,2,2,2,1,8,6,7,5,4,4,3,1,2,3). Since the die is fair, each of these outcomes are equally likely so we can just count how many of these outcomes have the desired number of 2's, 3's, etc.
There are $8^{15}$ possible rolls that could occur. There are ${15 \choose 2}$ ways to choose which positions the twos could occur in, ${13 \choose 3}$ ways to choose the positions the threes could occur in, …
In total this yields the probability:
$$\frac{{15 \choose 2}{13 \choose 3}{9 \choose 4}{5 \choose 5}}{8^{15}}$$