Count the number of doubles created from rolling three dice I have doubts about my calculations for the number of doubles created from rolling three dice. By doubles, I mean the outcomes of the form $aab$ where $a$ and $b$ are distinct values from $1$ to $6$.
In the case of two dice where I calculate the number of singles (outcome of the form $ab$). I can calculate it like this: $\binom{6}{2}\cdot2!=30$ (number of ways to choose two values from the set $1, 2, ..., 6$ times the number of arrangements of $ab$).
On the other hand, if I try to calculate the number of doubles created from rolling three dice, I get the incorrect result using the same logic for calculating number of singles from two dice: $\binom{6}{2}\cdot\frac{3!}{2!}=45$ (number of ways to choose three values from the set $1, 2, ..., 6$ times the number of arrangements of $aab$). It seems like I need to multiply by $2$ to get the correct result: $90$. I read other answers that $2$ represents the number of ways to choose a value for a pair but I don't understand why we need to multiply by $2$ here. I need an intuitive explanation on this part.
Related question: Why this is true?
$$\binom{6}{1}\binom{5}{1}=\binom{6}{2}2$$
Why there is need to multiply $\binom{6}{2}$ by $2$?
 A: You wish to find the number of outcomes in which one number appears twice while another number appears once when three six-sided dice are rolled.
There are six ways to select the value that appears twice, $\binom{3}{2}$ ways to select the two dice on which that number appears, and five ways to select the value that appears on the other die.  Hence, there are
$$6 \cdot \binom{3}{2} \cdot 5$$
such outcomes.
As for your approach, there are $\binom{6}{2}$ ways to select which two values appear, two ways to select which value appears twice (you omitted this step), and $\binom{3}{2} = \frac{3!}{2!1!}$ ways to distribute the values on the two dice, so there are
$$\binom{6}{2}\binom{2}{1}\binom{3}{2}$$
such outcomes.
As for the problem, you posed in the comments about finding the number of outcomes in which exactly one value appears twice and three other values each appear once when a die is rolled five times, choose which of the six values will appear twice, choose on which two of the five dice that value will appear, choose which three of the five remaining values will each appear once, and arrange those three distinct values on the remaining three dice.

 $$\binom{6}{1}\binom{5}{2}\binom{5}{3}3!$$

Alternatively, choose which four of the six values will appear on the five dice, choose which of these four values will appear twice, choose on which two of the five dice that value will appear, and arrange the remaining values on the remaining three dice.

 $$\binom{6}{4}\binom{4}{1}\binom{5}{2}3!$$

A: You want the $\binom{6}{2}2!$ possibilities of ab times the $3$ possibilities for aab.
