How many ways are there to put 5 balls in 2 boxes if the balls are distinguishable and the boxes are indistinguishable?

Guest Aug 4, 2022

#1**+1 **

Consider it case-by-case.

Case 1: 5 balls go in 1 box, and none go in the other.

There is \({5 \choose 5} = 1 \) way to do this

Case 2: 4 balls go in 1 box, and 1 goes in the other.

There are \({5 \choose 1} = 5\) ways to choose the balls to go in 1 box, leaving the other 4 for the other box.

Case 3: 3 balls go in 1 box, and 2 go in the other.

There are \({5 \choose 3} = 10\) ways to choose the balls to go in 1 box, leaving the other 2 for the other box.

So, there are \(1 + 5 + 10 = \color{brown}\boxed{16}\) ways.

BuilderBoi Aug 4, 2022

#2**0 **

Here's another way:

Each ball has 2 choices to go into a box. There are 5 boxes, so there are \(2^5 = 32\) combinations.

But, because the boxes are indistinguishable, we divide by 2. (Putting balls A, B, and C in 1 and D and E in the other is the same putting D and E in 1, and A, B, and C in the other)

So, there are \(32 \div 2 = \color{brown}\boxed{16}\)

BuilderBoi Aug 5, 2022