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I want to know how many combinations of three dice given as a product an even number. If these dice are different from each other the answer would be $6^3- 3^3$. What about if these dice are equal to one another? Like having $2, 4, 6$ is the same as $4, 6, 2$.

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    $\begingroup$ Do you know how to enumerate the number of possible results of three dice under the constraint? I.e., what takes the role of '$6^3$' if you're talking about the set of dice and not the individual dice? $\endgroup$ – Steven Stadnicki Aug 19 '19 at 22:12
  • $\begingroup$ @StevenStadnicki could you be more specific? $\endgroup$ – user42912 Aug 19 '19 at 22:23
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Have you heard of the stars and bars method? In this case, each die would be represented by a star. Any dice before the first bar is a 1, any between the first and second bar is a 2, etc. So, for example, |*||**|| would represent one 2 and two 4's. There are 8 possible locations to choose 3 stars, so the total number of results from rolling 3 dice is $\binom83=56$.

You can follow similar logic to count the results where all dice are odd.

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First you need to know how many combinations you can get from three dice if the order doesn't matter. This is what Steven Stadnicki meant in his comment.

There are $6$ ways if all three numbers are the same. If two of the numbers are the same, and the other is different, there are $6\cdot5=30$ ways. If all numbers are different, there are ${6\choose3}=20$ ways. Altogether that makes $56$ ways.

Now do the same kind of analysis to count the cases where all the dice show odd numbers.

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