How many ways are there to put 5 balls in 2 boxes if the balls are distinguishable and the boxes are indistinguishable?
+2666 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.
+2666 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 \(32 \div 2 = \color{brown}\boxed{16}\)
