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So what I think is the way to solve this question is to first count the total number of ways of putting all the balls into boxes such that the restriction isn't satisfied (i.e. the total number of ways of putting the balls into the boxes). Then using the inclusion-exclusion principle for situations where the boxes are empty.

I think the number of total ways that you can distribute them would be $4^5$ but I'm unsure how to do the second part of the question.

IdkWhyIdk
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  • In order to understand a general solution, you will have to look at P.I.E or Stirling Number of the second kind. For this specific problem, given there is just one extra ball than the bin, you can have a much simpler solution as provided in below answers. – Math Lover Nov 15 '20 at 15:57
  • Hi, I have done a little bit of reading on a more general solution and I'd forgot to mention I wanted to solve this type of question using a general method in the question. So it's my fault for being so forgetful – IdkWhyIdk Nov 15 '20 at 16:00

3 Answers3

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Your first part is correct. The number of the total ways is $4^5$.

Then you might consider the in-exclusion principle: $A_1$: $Box_1$ is empty; $A_2$: $Box_2$ is empty; $A_3$: $Box_3$ is empty; $A_4$: $Box_4$ is empty.

Use the same method that you got the number of total ways. For each of these cases, distributing $5$ distinguishable balls to $3$ distinguishable boxes, number of ways is $3^5$. $\binom{4}{1}$ shows it has $4$ cases.

$\vert A_1 \bigcap A_2\vert$ (Both Box 1 and Box 2 are empty) $= 2^5$. $\binom{4}{2}$ (from $A_1$ - $A_4$, choose $2$) shows how many cases in this situation.

$\vert A_1 \bigcap A_2 \bigcap A_3\vert$ (Boxes 1,2,3 are empty) $= 1^5$. $\binom{4}{3}$ cases.

$\vert A_1 \bigcap A_2 \bigcap A_3 \bigcap A_4 \vert$ (Boxes 1,2,3,4 are empty)= $0^5$. $\binom{4}{4}$ case.

Then apply inclusion-exclusion principle, $\vert A_1 \bigcup A_2 \bigcup A_3 \bigcup A_4 \vert$(at least one box is empty) $$=\sum_{k=1}^4 (-1)^{k+1}(4-k)^5 \binom{4}{k}$$

Finally, use the total number $4^5$ minus the number we got from the last step, then we get the number of ways that no box is empty.

To be honest, this method is much more complicated than other approaches. However, I guess you might want to generalise it to distribute $r$ distinguishable balls into $n$ distinguishable boxes(?) Then it definitely should be a good way to think about this question.

N. F. Taussig
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WoGenFaker55Kai
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  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. I placed the $-1$ in parentheses to account for the precedence of exponentiation over multiplication in the order of operations. – N. F. Taussig Nov 16 '20 at 00:35
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One box will have two balls. Choose which box that will be, choose two balls to put in it, then put one ball in each of the remaining boxes.

$$\binom{4}{1}\binom{5}{2}\times 3!=4\times 10\times 6=240$$

Daniel Mathias
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Choose which two balls are to be paired, $\binom{5}{2}=10$, and then arrange the four objects, $4!=24$, for a total of $240$.

JMP
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