Two identical dice are thrown simultaneously. Find probability of getting a $3$ and a $2$. 
Two identical dice are thrown simultaneously. Find the probability of getting a $3$ on
  one of the dice while a $2$ on the other.

I'm pretty sure the answer is $\frac{2}{36}$ but my friend says the answer should be $\frac{1}{21}$.
He argues that as the dice are identical, so sample space comprises only $21$ combinations.
I'm unable to explain it to him but I don't agree with his answer of $\frac{1}{21}$.
Can someone tell me the correct answer and provide a short explanation?
 A: There might be 21 possible outcomes ( (1,1), (1,2),....(6,6) ) , but they don't have equal probability, 1+1 can only occur in one way, but 2+3 can occur two different ways.  There are 6 ways to throw (1,1), (2,2), (3,3), etc, and 15 different pairs (1,2)/(2,1), (1,3)/(3,1), etc, making 36 possible combinations. We can get a 2 and 3 in two different ways, so the probability is 2/36 as you say. 
A: There are 36 possible outcomes. If it doesn't matter the order of the die, then the probability is $2* (1/36)$ ways it can happen. If it has to be 2 first, then 3 next, then it is $1/36$.
A: The odds of getting a 3 on a die is 1/6
The odds of getting a 2 on a die is 1/6
There are two possibilities for a 3 on one and a 2 on the other:
1. Die #1 is 3 and Die #2 is 2
2. Die #1 is 2 and Die #2 is 3
2 possibilities * 1 in 6 * 1 in 6 = 2/6/6=1/18 (or 2/36)
A: As the outcomes of dices should be $2$,$3$. So $2$ on one dices and $3$ on another as dices are identical we can't distinguish them so only one way that is $(2,3)$. Now, the sample space for this will be $21$ not $36$ because the dices are identical, i.e. $(1,2)$ is not different from $(2,1)$ etc.
So sample space = $21$ and probability = $1/21$.
