Dice rolled 6 times. What is the probability to roll 2 twice, 4 twice, and 6 twice? Also, what is the probability to roll 2 thrice, and roll 4 thrice?

Question

Here's what I currently have, however im not sure its correct.

a) $\mathbb{P}(2 \text{ twice}, 4 \text{ twice}, 6 \text{ twice}) = \{a, a, b, b, c ,c\}$ where $a$, $b$, $c$ are numbers between $1$ and $6$.

Say $a = 2$, $b = 4$, and $c = 6$. The order of the rolls don't matter and so we can order $6$ elements in $6$ fac ways. Therefore the probability $= \frac{6!}{6^6} = \frac{720}{46656}$

b) The same result, simply changing the set above with ${a, a, a, b, b, b}$ with $a = 2$ and $b = 4$.

Is this correct?

Answer 1

a) Taking $6!$ as numerator you are overcounting. In stead you should take: $$\binom6{2,2,2}=\frac{6!}{2!2!2!}=90$$

b) Here as numerator you should take: $$\binom63=\frac{6!}{3!3!}=20$$

To get some insight try to count how many sixtuples there are containing $2$ and $4$ both three times.