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 $2$?

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