Another clever trick
We have 4 ways to choose D, 3 ways to choose C, to get 4 times 3 = 12 ways. Now we need to consider the restriction that the problem gave us. A has to be to the left of B. After we choose D and C already, we know there are only 2 spots left to place A and B, so there would only be one way to
place A and B after placing C and D. Thus, our answer is 4 times 3 times 1 = 12 ways.
@professionalbronco Yeah that's exactly right! Great job for coming up with that, it's a really crucial concept for counting problems.
Actually, if you notice, towards the end of that video in Part 5 Prof. Loh does something very similar, he divides by 2 to account for each "pair" since only one arrangement in each fulfills the conditions. It's basically what you were thinking of but kinda the reverse! Great minds think alike