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Part (a) How many ways are there to put $4$ balls into $3$ boxes, given that the balls are not distinguished and neither are the boxes?

Part (b) How many ways are there to put $2$ white balls and $2$ black balls into $3$ boxes, given that balls of the same color are indistinguishable, but the boxes are distinguishable?

I am not sure how distinguished and indistinguishable makes a difference in the question and I have no clue how to do this. Any help is appreciated.

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For part (a), these are (unordered) partitions of $4$ into $3$ parts where parts can be $0.$ To ensure that you hit all possibilities, tackle three cases: exactly $2,1,$ or $0$ parts with nothing. There cannot be three parts with nothing. If exactly $2$ parts have nothing, the only possibility is $4+0+0.$ If exactly $1$ part has nothing, the possibilities are $3+1+0$ and $2+2+0.$ If exactly $0$ parts have nothing, then each part has at least $1.$ This means the remaining one gets assigned somewhere and the only possibility is $2+1+1.$ So the answer is $4.$

For the second part, treat white balls and black balls separately. Doing either one will let you square it to get the answer thanks to the multiplication principle. For white balls, there are $$\binom{2+3-1}{3-1}=\binom{4}{2}=6$$ possibilities, so the answer is $6^2=36.$ This is using sticks and stones.

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    $\begingroup$ I believe part (b) is wrong. The question states that there are $3$ boxes, and I think you are considering only $2$. $\endgroup$
    – Bergson
    Jun 23, 2020 at 18:26
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    $\begingroup$ @ThomasBladt thanks, you are right that I misread $3$ boxes for $2$ boxes. I have fixed the solution. $\endgroup$
    – Favst
    Jun 23, 2020 at 18:30

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