If a Ball can be any of $n$ colors, then the number of configurations (with repetition of colors) of $k$ balls is $(n+k-1)C(k)$ why?

Question

For example, if a ball can be any of 3 colors, then the number of configurations (with repetition of colors) of 2 balls is $(3+2-1)C_{2} = 4C_{2} = 6$ Why?

Answer 1

i found the answer from the wiki page

https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

and the video

https://www.youtube.com/watch?v=UTCScjoPymA

Answer 2

Configuration isn't a technical term, but you must be talking combinations and not permutations. Otherwise the comments would be correct. So we're talking in totals. if there are 10 balls and red/white for colors, you could have 10 red, 9 red 1 white, 8 red 2 white ... 0 red 10 white. That's 11 ways it could be. In your formula, it would be 2+10-1C10 or 11C10 = 11. So the formula checks out in that example.

In general, You're selecting how many balls are each color. In order to keep track of all the ways it could happen, we can organize it like this: Suppose there is a bucket of each color of balls and a clown standing in front of the first bucket. You can tell him two commands: "Ball" and "Next". If you tell him "Ball" he will give you a ball from the bucket. If you tell him "Next" he will move to the next bucket. As long as you say "Ball" K times, you will end up with K balls. As long as you say "Next" N-1 times, you will have had a chance at every color. It's possible to get any arrangement of colors you want, so it's complete, and every different order of commands will give you a different count of each color so it's unique.

This turns it into an easier counting problem: How many different ways can we arrange K ball commands (B) and N-1 next commands (N)? There will be K+N-1 total commands and we need to choose K of them to be B. So (K+N-1)C(K)