In how many ways can you distribute $8$ indistinguishable balls among $5$ distinguishable boxes, if at least three of the boxes must be empty?

RedDragonl May 30, 2024

#1**+1 **

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! :)

NotThatSmart May 30, 2024