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Say we have 5 distinct balls being placed into two distinct bins. If each ball has 2 options of where it can go, This would make (2x2x2x2x2) combinations, 2^5.

If each bin may not be empty, how would we account for and subtract those cases where a bin is left empty using combinatorics or logic?

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    Hint: How many ways are there to have one or other of the bins empty? – 2017-01-26

2 Answers 2

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One can easily use the complement-rule here. Let $A$ denote the set of all possible arrangements. Let $A_1$ and $A_2$ $A_{1,2}$ denote the events that (respectively) bin 1 and 2 are empty. It should be clear that: $A= A_1 \cup A_2 \cup (A_1 \cup A_2)^c.$

It then follows that (because $A_1$, $A_2$ and $(A_1 \cup A_2)^c$ are disjunct and finitely countable):

$A= \vert A_1 \vert + \vert A_2\vert + \vert(A_1 \cup A_2)^c\vert$.

So it simply says, that the number of configurations without empty bins is all the configurations minus the (two) possible configurations with one empty bin (as it is impossible to have them both empty).

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    That makes sense. Brief followup: Could this same method be applied if the numbers were changed to 4 balls and 3 bins? I'd assume the formation of sets would be slightly more complicated. – 2017-01-26
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    It can be yes! Yet, when having three bins, it is possible to have two empty bins, so that gives you a lot more options. – 2017-01-26
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    Complement rule you add in your answer doesn't make any sense. – 2017-01-26
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If all balls are placed in 1st bin. In only $1^5 = 1$ way.

Similarly all balls are placed in 2nd bin.

So we have 2 ways either 1 or 2 bin is empty.

Cases with any bin can not be empty in

Total cases - Cases of empty bins =

$2^5 - 2$

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    Thank you. If it's not too much trouble, would you be able to briefly explain a similar scenario in which there are 5 balls and 3 bins instead? It seems to become a bit more complex when there are 3 bins and I am trying to avoid double counting, starting with 3^5 total combinations and removing the invalid ones. – 2017-01-26
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    Why not you post a new question. It seems interesting question. And I also try to solve it. After asking send me the link. – 2017-01-26
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    http://math.stackexchange.com/questions/2115184/placement-of-unique-objects-into-distinct-containers – 2017-01-26