Yet another way
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The way I did it, and I thought he would too, was by considering the possible placements of B for any place of A. Since after we have defined 2 of the 4 elements, the other two can only be arranged in two ways. Then, you can add up all the ways to put A before B with considering the other two elements as one singular one, and then multiply by 2.
So, with A in front 3 placements work, second, 2 and so on, so that we get 3+2+1=6. Then, we multiply by 2 to get 12.
this also works on the your turn, except that there is actually only 1 arrangement of C and D for every placement of b. -
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