# Possible combinations for rolling 5 similar dice

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?

Thanks,

## 1 Answer

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.