What is the total possible combinations for an two dice throw. Why is the number of combinations for two dice throws $6\times6$ which is $36$ but not also ${12}\choose{2}$ divided by two which is $33$?  This corresponds to choosing two elements from the lot which is $6$ possible numbers from every dice, meaning $12$ in total and dividing by two because the lot has every element repeated one time if I understand it right. Can you please explain what is missed in ${{12}\choose{2}} \cdot \frac{1}{2}$ or why my logic is wrong to consider ${{12}\choose{2}} \cdot \frac{1}{2}$ as the answer?
 A: Say we have two dice labeled $a$ and $b$. The $12$ outcomes are,
$\{a1, a2, a3, a4, a5, a6, b1, b2, b3, b4, b5, b6\}$
You are counting $ \displaystyle {12 \choose 2} = 66$ outcomes. Now for example, out of $ \{a1, a2 \}$ and $ \{a1, b2 \}$, only the latter is valid. But you cannot simply divide by $2$ as there is no $ \{a1, a1\}$ for example. So you first need to subtract $6$ when both numbers are same. That gives you $60$ outcomes and out of that, only $30$ are valid. That leads to $30 + 6 = 36$ outcomes.
A: In using $\frac{_{12}C_2}{2}$ to calculate the number of possible combinations, you assume that each possible dice roll appears twice in the two-element subsets of the 12-element set containing each face twice. This is true for when you roll two different numbers (one die has the higher number, other die has the higher number). But when the two numbers are the same (ie. you roll doubles), they are only counted once, so your assumption is false.
Since the two rolls are independent, it's better to consider the possibilities separately and multiply them together:
$$N=\binom{6}{1}\binom{6}{1}=6\cdot6=36$$
If you wanted, you could calculate the number of possibilities by breaking it down into cases. If you roll doubles, there are clearly $\binom{6}{1}$ ways to do that. If you don't roll doubles, then you have two different faces, with $\binom{6}{2}$ options. Then each of those can be ordered in 2 ways, so:
$$N=2\binom{6}{2}+\binom{6}{1}=2(15)+6=36$$
A: $(1,1),    (1,2),    (1,3),    (1,4),    (1,5),    (1,6)$
$(2,2),    (2,3),    (2,4),    (2,5),    (2,6)$
$(3,3),    (3,4),    (3,5),    (3,6)$
$(4,4),    (4,5),    (4,6)$
$(5,5),    (5,6)$
$(6,6)$
Throwing two regular dice, the maximum number of combinations is 21 and no more.
See the combinations above.
