+202 Find the number of ways of distributing 4 different balls among 4 identical boxes.
So I was thinking that it would be \(4^4/4\) since you have 4 choices for each ball but there are 4 we since the boxes are identical we're overcounting by a factor of 4. Is this right?
+118667 Find the number of ways of distributing 4 different balls among 4 identical boxes.
Counting questions are always tricky, but here is my thinking.
| distribution of balls | ||
| 1,1,1,1 | 1 way | |
| 2,1,1,0 | 4C2=6 ways | |
| 2,2,0,0 | 4C2/2=3 ways | |
| 3,1,0,0 | 4C3=4 ways | |
| 4,0,0,0 | 1 way | |
| Total = 15 ways |
So no, I do not think your way makes any sense.
Feel free to ask questions.
The red is my edit. Impasta found a fault. Good logic Impasta!
+202 So are you implying for 2-1-1-1, for example, that out of the four balls, we choose 2 out of four balls to be put in one box, and we don't need to account for permutations?
+202 Wait, I'm pretty sure doing it that way overcounts some things. For instance, doing \(\binom{4}{2}\) for 2-2-0-0 will count each way twice since you choosing two balls or you choosing the other two will result in the same balls in boxes. But building off of your idea, I think I can get the answer. Thanks!
