My question is related to this question. We throw five ordinary six-sided dice and tries to obtain various combinations. Specifically I want to know about the case where Two dice showing one number, two other dice showing a different number, and the fifth die showing a third number. The number of possible outcomes for this case, that is 2-2-1, is $\binom62\binom41\times \frac{5!}{2!2!} = 1800$. But my question is why?

The way I am currently thinking about is that, we first need to choose 3 different numbers: $6 \times 5 \times 4$. Now, we have three classes of sizes 2,2, and 1. So, using multinomial coefficient we get: $\frac{5!}{2!2!1!}$. If we follow this thought, we get: $6 \times 5 \times 4 \times \frac{5!}{2!2!} = 3600$. What's wrong here?



The order in which you choose the two numbers that are repeated doesn't matter. For example, if you choose $(11)(22)(3)$ or $(22)(11)(3)$, those are actually the same, but you've counted them twice in your calculation.

So you'd need to divide by $2$ to compensate for the overcounting, which gives you the correct result.


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