Say 6 people are choosing between 9 fruits on a menu: apple (A), banana (B), cherries (C), dates (D), elderberry (E), fig (F), grapefruit (G), honeydew (H), Indian prune (I).
If each person has an equal likelihood of choosing each of the 9 options, what's the probability of having 1, 2, 3, 4, 5, or 6 distinct fruits chosen? So for instance, if the 6 people choose the following fruits - apple, banana, apple, apple, grapefruit, apple - that's 3 distinct fruits.
For 1, I can calculate it because they are independent choices. There are 9 possible fruits to be selected, and each person has 9 choices. It's 9/9^6. For 6, I can calculate as well. That's (9*8*7*6*5*4)/9^6.
I can't figure out how to theoretically calculate any numbers in between 1 and 6, i.e. 2, 3, 4, or 5. Through simulation, I know what these should be (2 = ~0.004, 3 = ~0.08, 4 = ~0.37, and 5 = ~0.42).
The total number of choices is $9^6$, so you need to find the number of "good" choices, i.e. ones where exactly $x$ out of $9$ options are chosen. Split this procedure into 2 steps. First, choose $1\le x\le 6$ particular options in $\binom{9}{x}$ ways. Then you just need to find the number of surjective (i.e. onto) functions from the set of 6 distinguishable elements (the people) to the set of $x$ distinguishable elements (the options). This is part of any discussion of The Twelvefold Way, and is equal in this case to $x!S(6,x)$, where $S(n,k)$ is the Stirling number of the second kind. So, the number of "good" choices is the product of the numbers of choices at each of the two steps, i.e. $\binom{9}{x}x!S(6,x)$, and the probability you're looking for is therefore $$ \frac{\binom{9}{x}x!S(6,x)}{9^6}. $$