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

 Aug 4, 2022
 #1
avatar+2437 
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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. 

 Aug 4, 2022
 #2
avatar+2437 
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 \(32 \div 2 = \color{brown}\boxed{16}\)

 Aug 5, 2022

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