If you roll a die two times, what is the probability the sum of the upturned faces equals $7$? If you roll a die two times, what is the probability the sum of the upturned faces equals $7$?
I can answer this question if I consider the order of the rolled numbers relevant. However, when I disregard the order I get a wrong answer.
My reasoning: we roll the die two times and each roll there are $6$ possible outcomes, so we have $$\binom{2+6-1}{2}=21$$ possible outcomes. There are three scenarios in which the sum of the upturned faces equals 7, namely: $\{1,6\}$, $\{2,5\}$ and $\{3,4\}$. That means the probability the sum of the upturned faces equals $7$ is $\frac{3}{21}=\frac{1}{7}$. It should be $\frac{1}{6}$. 
Where am I going wrong?
 A: Since the two die rolls are independent, it should come as no surprise that there are $6\cdot 6 = 36$ possible pairs of die rolls.
Of those rolls, only the following pairs have a sum of $7$:
$$(1,6),\ (2,5),\ (3, 4),\ (4, 3),\ (5, 2),\ (6, 1)$$
Naturally, then, the likelihood of our two die rolls summing to $7$ would be:
$$\frac 6{36} = \frac 16 \approx 16.6\%$$
A: We roll the die two times, and each roll there are $6$ possible outcomes, so we have $6^2=36$ possible outcomes. If you disregard the order, you must count each favorable result $2!=2$ times. Your result then will be $\frac{3\cdot2!}{36}=\frac{6}{36}=\frac{1}{6}$ (Note: this logic will fail when the favorable results are not equally probable.)
A: There is, in fact, nothing to prevent you from defining a distribution over all ordered pairs of integers $(x,y)$ such that $1 \leq x \leq y \leq 6,$
which covers exactly $21$ possible pairs,
and there is nothing to prevent you from assigning probability $\frac1{21}$
to each of them.
But what does that have to do with rolling a die?
One nice thing about dice problems is that real-life dice have well-known behaviors that are a very good match for the mathematical model of dice.
Each side is equally likely to come up (as far as you're every likely to determine, at least), there are no obvious ways in which the number showing on one die would be systematically influenced by the number showing on another (or by the previous roll of the same die), and so forth.
Real-life dice also have this very important property:
They don't care how you count.
If you repeat the experiment (rolling a die twice) many times,
someone watching you and counting how many times each number came up on the first roll will probably see each number come up about $\frac16$ of the time.
And among all the pairs of rolls in which the first roll was a $1,$
each of the numbers will come up on the second roll about $\frac16$ of the time. That's because each side is equally likely to come up and the outcome of one roll doesn't affect the outcome of another roll.
As a result, $(1,1)$ comes up in $\frac16$ of $\frac16$ (that is, $\frac1{36}$) of the observations,
$(1,2)$ comes up in another $\frac1{36}$ of the observations, 
$(2,1)$ in another $\frac1{36}$ of the observations, and so forth.
Since $(1,2)$ and $(2,1)$ are the only ways to get a sum of $3,$
and $(1,1)$ is the only way to get a sum of $2,$ this observer will see a sum of $3$ about twice as often as a sum of $2.$
If you don't pay attention to the order in which the numbers in each pair of rolls came up, but consider any pair of rolls with a $1$ and a $2$ as the same outcome, will you see a higher ratio of sums of $2$ to sums of $3$ than the other observer? How can that be?
Now suppose this observer goes away and you continue rolling the die,
but you count any pair of rolls with a $1$ and a $2$ as the same outcome.
Will the dice suddenly start coming up with a total of $2$ as often as the total is $3$?
It seems unreasonable to expect the die to change its behavior just because you record the results differently than someone else would.
Since we know a very clear and understandable mechanism by which sums of $3$
come up twice as often as sums of $2$ on two consecutive rolls,
and we have no physical model (at least, I've never heard of one) that would explain how those two sums could come up equally often,
we consider the sum $3$ to be twice as likely as $2$ no matter how you write down the numbers of observations (recording whether the $1$ came before the $2,$ or not recording that detail).
So you're welcome to partition your sample space so that the sequences of rolls $(1,2)$ and $(2,1)$ are not distinguished, as long as you either (a) make the probability of a $1$ and a $2$ twice the probability of two $1$s
or (b) don't claim to be talking about two rolls of a fair die.
