2021 AMC 10A Prob 18

💚 Better Counting Method I found a simpler method after a second look. Wlog, assume the upper left box has crop "1" and then break into four subset cases based on the choice of the lower right crop: 1 x → 3 choices for each x x 1 3 • 3 = 9 (diag same) 1 x → 2 choices for each x x 2 2 • 2 = 4 (diag diff) 1 x → 2 choices for each x x 3 2 • 2 = 4 (diag diff) 1 x → 2 choices for each x x 4 2 • 2 = 4 (diag diff) 9 + 4 + 4 + 4 = 21, but the assumed upper left box could have been 1, 2, 3 or 4, so the total count is 4•21=84 ■ A useful counting strategy is to identify choices that are independent and push those to front of the counting process. For this problem, the two crops across a diagonal are independent choices IF those two choices are made first. 🤔 Tough Counting Problems I had an uneasy feeling that my first method was not the most efficient, but it would probably get me to an answer in about 5 minutes. How long might it take to discover a better method? Probably longer than 5 minutes! For the AMC in general, if you find an approach that likely gets you an answer within 5 minutes, then just stick with that approach. ♦ 2 Types → Opposite Group I could have counted this subset more efficiently. The two crops do NOT have a neighbor restriction, so we have a simple binary choice (eg. 1 or 3) for each of the four squares → 2•2•2•2 = 16 arrangements, but this includes 1111 and 3333 that don't belong in this subset, giving 16-2 = 14 arrangements for any distinct pair of opposite group crops.