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Would it be as simple as $\displaystyle {6 \choose 2}\times{5 \choose 3}\times{5 \choose 2}$?

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    Are the two numbers specified? The number of ways to get three $1$s and two $2$s is not the same as the number of ways to get three of one number (which could be any number) and two of some other number. – 2017-02-04

2 Answers 2

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First pick any three particular dice: $\dbinom{5}{3}$

Select any one number now which appears in the selected three dice $\dbinom{6}{1}$

Now pick remaining two dice: $\dbinom{2}{2}$

Select any of the remaining number which appears in the selected two dice $\dbinom{5}{1}$

So, number of ways is

$\dbinom{5}{3}\dbinom{6}{1}\dbinom{2}{2}\dbinom{5}{1}$

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    What's wrong with the answer $nCr(6,3)$ times $nCr(6,2)$ ? I got the same answer as you this way. – 2017-02-04
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    how u got nCr(6,3) times nCr(6,2)? – 2017-02-04
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You're on the right track, but there are a few misses. First off, you're not choosing $2$ values freely, you're choosing one value of which to have three dice, and then one value of which to have two. Therefore it's $6\cdot 5$ instead of $\binom 62$.

Second, you only need to choose which of the five dice are to be three together, and the two remaining will be chosen automatically. Therefore it's $\binom 53$ instead of $\binom53\binom52$.

So the end result is $6\cdot 5\cdot\binom53$.