In how many ways can you distribute $8$ indistinguishable balls among $5$ distinguishable boxes, if at least three of the boxes must be empty?
We can use casework to solve this problem.
If 3 boxes are empty, then there are two possibilites.
First, we could have 2 boxes with balls in them.
Case 1
Let's say one box and has 7 and the other has 1.
There are 2 ways we can do this.
Case 2
Now, let's say there is one boy with 6 and one box with 2,
There are also 2 ways we can do this.
Case 3
One box has 3 and the other has 5.
Another 2 ways we can do this.
Case 4
One box has 4 and the other has 4.
Only one way to do this.
\(2 * 2 * 2 * 1 = 8 \) ways we can do this in total
We choose 2 boxes out of the 5 we could have chosen from, which is just \(5 \choose 2\) = 10.
\(10 * 8 = 80\) ways to do 2 boxes.
Now, let's say only 1 box has all the balls. There are 5 ways to choose a box with all 8 balls in it.
\(80+5=85\). There are 85 ways we could do this!
Thanks! :)