An Ace has a value of 11 in this problem, and a face card has a value of 10. Thus, my idea for solving this problem was to separate the problem into 7 different cases (drawing a 2 first, 3 first, 4 first, etc.), because drawing anything greater than an 8 will result in a sum greater than 11 no matter what.
So for the case of drawing a 2 first, the number of ways was 4 x ((6x4) + 3), because 6 possible numbers could be drawn plus the three 2's that were left to add up to less than 11. For the case of 3, it would be 4 x ((5x4) + 3). I think I'm on the right track, but I'm unsure of whether or not I would multiply each case by 2! to account for the order of cards drawn not mattering?
Any help would be appreciated, or even a different solution to this problem :)