So I was learning about coin/die tossing in classes and that got me thinking about something. If we take k n-sided die then the total number of outcomes is n^k. But that assumes that order matters (each roll is significant). How do we figure out the number of significant outcomes.
For eg. in a 6 sided die rolled twice: (1,2) and (2,1) are the same and only should count as a significant roll. For k=2, I've determined that the number is n(n+1)/2
Have you heard of stars and bars? Each die would be represented by a star while the bars are separators. Everything before the first bar is a 1, everything between the first and second bars is a 2, etc. For example, $*|**||*|**|$ would represent one 1, two 2's, one 4, and two 5's. We need n-1 bars to separate out n possibilities for each n-sided die and k stars for k dice. The number of different outcomes is the number of ways we can choose the positions of k stars out of k+n-1 possible positions. So the number of possible outcomes is $\binom{n+k-1}k$.