Find the number of ways of distributing 4 different balls among 4 identical boxes.

I'm really stuck on this one.

Guest Feb 9, 2020

#2**0 **

*Find the number of ways of distributing 4 different balls among 4 identical boxes.*

You have to make a choice which ball to put in the first box. You have 4 options.

Then you must make a choice which ball to put in the second box. You have 3 options (there are three balls left)

Then you have to make a choice which ball to put in the third box You have 2 options (there are two balls left)

It isn't necessary to make a choice about the fourth box, because there's only 1 option (there's only one ball left)

So you have 4 choices for the first box.

For each of those 4 choices, you have 3 choices for the second box. That makes 12 number of ways.

For each of those 12 ways, you have 2 choices left for the third box. That makes 24 number of ways.

You have only 1 ball left for the fourth box, so that leaves you at ................. ....... **24 number of ways**.

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Guest Feb 9, 2020

#5**0 **

After I put the answer in, this was the answer it gave me. Turns out it was actually 15. Thank you still for trying to help! Much appreciated!

Case 4-0-0-0: There is only one way.

Case 3-1-0-0: There are four ways to choose the ball that is by itself. Then the other three balls must go in a different box.

Case 2-2-0-0: There C(4,2) = 6 ways to choose two balls to go in one box. Then the other two balls must go in a different box. But since order doesn't matter (there is no first box or second box), we divide by 2.

Case 2-1-1-0: There are C(4,2) = 6 ways to choose two balls to go in one box. Then the other two balls go in separate boxes.

Case 1-1-1-1: There is only one way.

This gives a total of 15 ways.

Guest Feb 9, 2020

#3**+1 **

Well, we know we can split the balls into four groups: 4-0-0-0, 3-1-0-0, 2-2-0-0, 2-1-1-0, 1-1-1-1.

Case 1: 4-0-0-0 has only 1 way we can assign the balls to different boxes.

Case 2: 3-1-0-0 has (I think) 4 ways we can assign the balls to different boxes. The one ball can go anywhere, and the three other balls must go in a different box

Case 3: 2-2-0-0 has 4C2 = 6 ways to choose two balls to go in one box, then the other two balls must go in separate boxes. But since order doesn't matter, (there is no first box or second box), we divide by 2.

Case 4: 2-1-1-0 has (again) 4C2 = 6 ways to choose two balls to go in one box, then the other balls go in seperate boxes.

Case 5: 1-1-1-1 has only 1 way.

**1 + 4 + 6/2 + 6 + 1 = 15.**

Please check for me.

AnimalMaster Feb 9, 2020