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I am confused

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How many ways can you distribute \$4\$ different balls among \$4\$ different boxes?

This is different than a problem on web2.0calc, that might seem familiar.

May 21, 2020

#1
+744
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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.

May 21, 2020
#2
+732
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Hi Sarvajit,

I think the answer is \(\boxed{256}\) because there are 4 choices for each ball, and each choice is completely independent. So, \(4^4=\boxed{256}\).

I'm not sure if this is right...

Please tell me if it is!

:)

May 21, 2020
#3
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OMG you are such a lifesaver... 256 was correct.

15 is not because my question is saying 4 different boxes, not 4 identical boxes @whymenotsmart, but thank you all for trying

May 21, 2020
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+744
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May 21, 2020
#5
+978
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also just to say this is an AoPS question.

Just sayin.

hugomimihu  May 21, 2020
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+744
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^m^......

May 21, 2020