My initial reaction is to say that the answer is $6^4$, since 4 dice can have 6 outcomes. In my train of thought, the first dice can have 6 outcomes, same as the second, third and fourth, thus $6*6*6*6$ would seem fitting. However, my professor claims that answer is wrong because it overcounts, and says that the answer is actually $9 \choose 4$.
Can someone help me understand how it overcounts and why the answer is $9 \choose 4$?
Best Answer
Your answer of $6^4$ would be correct if we were considering ordered outcomes, where it matters what each die shows in order. In this scenario, you'd treat $1,1,1,2$ as different from $1,2,1,1$ for instance.
Your professor is considering unordered outcomes, which would treat both of the examples above as $3$ ones plus $1$ two.
One way to count the unordered outcomes is a stars and bars approach - Wikipedia. Here you imagine your 4 dice as 'stars' and your $6$ individual outcomes (numbers $1\dots6$) as $5$ 'bars'. An outcome looks something like: $$\text{ones}\left|\vphantom\int\right. \text{twos}\left|\vphantom\int\right. \text{threes}\left|\vphantom\int\right. \text{fours}\left|\vphantom\int\right. \text{fives}\left|\vphantom\int\right. \text{sixes}$$
The combination $1,2,1,1$ (3 ones, 1 two) would look like: $$***\left|\vphantom\int\right. *\left|\vphantom\int\right. \left|\vphantom\int\right. \left|\vphantom\int\right. \left|\vphantom\int\right. $$
The combination $2,4,4,5$ (1 two, 2 fours, 1 five) would look like: $$\left|\vphantom\int\right. *\left|\vphantom\int\right. \left|\vphantom\int\right. **\left|\vphantom\int\right. *\left|\vphantom\int\right. $$
All together there are 9 objects (4 stars, 5 bars), and any combination amounts to picking 5 of the locations for the bars - $\dbinom95$ possibilities, or picking 4 locations for the stars - $\dbinom94$ possibilities. Happily, they give the same result.