Combinations with and without repetition: How many $6$ digit words you can assemble using each of the $0,1,2$ numbers twice? I am very confused with combinations. Here is a question:

How many 6 digit words you can assemble using each of the 0,1,2 numbers twice?

My attempt was to use the "with repetitions" formula (because the numbers appear twice, huh?):
$$D(n,k)={{n-1+k \choose k}}.$$
And the combination of all numbers should be: $D(6,2)⋅D(4,2)⋅D(2,2)=630$. So far, so good.
But I was shocked to hear that I am WRONG. Now, the right way to do that was using the distinct combinations:
$${{6 \choose 2}}⋅{{4 \choose 2}}⋅{{2 \choose 2}}=90.$$
May somebody explain or give a hint why we need to use the distinct combinations formula?
 A: You are counting positions into which to put digits.  (Two choices for positions of "$0$" out of six positions.  Then two choices for positions of "$1$" our of the remaining four positions.  Then no actual choice for the positions of the "$2$"s.)  Positions are not repeated, so you should not be using "with repetition".
A: Take two red marbles (representing zeros), two blue marbles (representing ones) and two white marbles (representing twos), and put six boxes in front of you.
To compose 'words', we have to put each marble into its own box. We start with the red marbles. For the first one we have six choices, and for the second one five. The order doesn't matter, as they are both red; this leaves ${6 \choose 2} = 15$ ways we can put the red marbles in the boxes. 
We now have 4 boxes left for our two blue marbles. Again, the first marble can be put in four boxes, and the second in three, and as order doesn't matter, we have ${4 \choose 2} = 6$ options.
The two white marbles go in the two remaining boxes. There were $15 \cdot 6 \cdot 1 = 90$ ways to arrange this.
A: How are you interpreting $D(n,k)$? There are a number of different interpretations, the most plausibly relevant one of which may be that $D(n,k)$ is the number of multisets (see note) of size $k$ that can be formed when there are $n$ elements to choose from. So when you write $D(6,2)$, you ought to be able to say what role multisets of size $2$ with $6$ elements to choose from play in your problem. If you think about this question for a bit, I think you'll realize that $D(6,2)$ doesn't make sense.
My guess is that you were somehow thinking about the two $0$s that your word needs to have. But if you're focussing on $0$, then there's only one element you're allowed to use, not six. So maybe $D(1,2)$ makes more sense than $D(6,2)$? But $D(1,2)=1$, and indeed, there's only one multiset of size $2$ that you can make using $0$, namely $\{0,0\}$. This isn't getting us very far, because talking about multisets doesn't say anything about the position of the digits in the word, and that's what matters in this problem.
One common approach to this kind of problem is to pretend initially that the two $0$s are distinguishable, and likewise for the two $1$s and the two $2$s. There are $6!$ ways to position the items $0_1$, $0_2$, $1_1$, $1_2$, $2_1$, $2_2$, where the subscripts allow us to distinguish the copies of a letter. If we now remove the subscripts so that, for example, the two words $0_11_11_22_20_22_1$ and $0_21_11_22_20_12_1$ become the same word, $011202$, then we need to divide by an "overcounting factor," $2\times2\times2=8$, in order to count only distinct words. This calculation matches up with
$$
\frac{6\cdot5}{2}\frac{4\cdot3}{2}\frac{2\cdot1}{2}=\binom{6}{2}\binom{4}{2}\binom{2}{2},
$$
which you mentioned is the correct answer, and whose interpretation was nicely explained in the other answers. Maybe you can think about how the two approaches to the problem match up.
Note: Multisets are sets with multiplicity, meaning elements may appear more than once. But, as with sets, multisets have no notion of order. So $\{0,0,1,1,1,2\}$ and $\{2,1,0,1,0,1\}$ are the same multiset. As an example of how $D(n,k)$ counts multisets, let's say we want to make two-element multisets using the elements $0$, $1$, $2$. There are six of these:
$$
\{0,0\},\ \{1,1\},\ \{2,2\},\ \{0,1\},\ \{0,2\},\ \{1,2\}.
$$
These correspond to the six ball-in-box configurations of JMoravitz's comment:
$$
**|\,|\,,\ |**|\,,\ |\,|**,\ *|*|\,\,,\ *|\,|*,\ |*|*,
$$
where the asterisks represent the balls, and the two bars separate the contents of the three boxes. And indeed, $D(3,2)=\binom{2+2}{2}=6$.
