Question 995681
If we put some distinguishing marks on the 2's,
we have 4 distinct items: {{{0}}} , {{{1}}} , {{{red(2)}}} , and {{{green(2)}}} .
The first digit cannot be {{{0}}} , because that would make the number less than 100.
(It would really be a 3-digit number in disguise, too).
So, we have 3 choices for the first digit.
For each of those choices, there will be 3 items left, to chose a 2nd digit from, for a total of {{{3*3=9}}} choices for the first two digits.
For each of those {{{9}}} choices, we still have to choose one of the {{{2}}} remaining elements as the third digit, giving us {{{2]]] choices.
That makes {{{3*3*2=18}}} sequences of 4 elements, where we can distinguish {{{red(2)}}} from {{{green(2)}}} .
However, taking those permutations of elements as numbers,
we would not distinguish between pairs with {{{red(2)}}} from {{{green(2)}}} in different orders,
like {{{10red(2)green(2)}}} and {{{10red(2)green(2)red(2)}}} ,
which both represent number {{{1022}}} .
So we need to divide by {{{2}}} , to find that there are
{{{3*3*2/2=highlight(9)}}} numbers greater than 1000 that can be made using the four digits of 2012.