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We have 20 objects, 10 of which are dice and 10 of which are balls - and we can distinguish them all - and we have two baskets; and in each basket we throw 10 objects. Can someone please explain to me how to calculate the number of possibilities of obtaining two baskets that contain the same amount of equals objects ?

In the solution to this problem, only "$\frac{10!}{(5!)^2} \cdot \frac{10!}{(5!)^2} $" was specified.

Would it make any difference, if we couldn't distinguish the objects ?

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For the first basket, we have to pick $5$ dice and $5$ balls. Since there are $10$ dice and $10$ balls, there are ${10 \choose 5}$ ways to choose the dice and ${10 \choose 5}$ ways to choose the balls, giving ${10 \choose 5} {10 \choose 5} = \frac{(10!)^2}{(5!)^4}$ choices in total. The other balls and dice you then simply put in the second basket.

If the objects are indistinguishable, then any two possibilities are the same: you simply have $5$ balls and $5$ dice in each basket. So then the answer would be $1$ possibility of putting the balls and dice in the baskets.