2021 AMC Fall 10A Problem 18

Each crop has exactly one "enemy" (corn–wheat, soybeans–potatoes). In the $2 \times 2$ grid, each section borders the two sections beside and below/above it but not the diagonally opposite one. So the two diagonal sections $P$ and $R$ are unrestricted with respect to each other, while each of the other two sections $Q$ and $S$ borders both $P$ and $R$.

Choose the crops of $P$ and $R$ first, then count options for $Q$ (and identically for $S$).

Case 1: $P$ and $R$ get the same crop ($4$ ways). Then $Q$ must avoid just one enemy, leaving $3$ choices; same for $S$. Total $4 \cdot 3 \cdot 3 = 36$.

Case 2: $P$ and $R$ get different crops ($4 \cdot 3 = 12$ ordered ways). Now $Q$ must avoid the enemy of $P$ and the enemy of $R$. These are two different crops (distinct crops have distinct enemies), so $Q$ has $4 - 2 = 2$ choices; same for $S$. Total $12 \cdot 2 \cdot 2 = 48$.

Adding: $36 + 48 = 84$.

The answer is $\boxed{\textbf{(C)}\ 84}$.

The adjacency structure of a $2 \times 2$ grid is a $4$-cycle, and the diagonal pair is the natural thing to fix: once those two are set, the remaining two cells are independent of each other and face identical constraints. Casework on "same or different" for the fixed pair keeps the count to two lines. In general, look for a set of mutually non-adjacent cells whose choice decouples the rest.

Group crops into families $\{$corn, wheat$\}$ and $\{$soybeans, potatoes$\}$. Adjacent cells from different families are always fine; adjacent cells from the same family must be identical. Assign families to the four cells, then pick one crop per connected block of same-family cells: all one family gives $2 \cdot 2 = 4$; three-and-one gives $8 \cdot 2^2 = 32$; two adjacent and two adjacent gives $4 \cdot 2^2 = 16$; two diagonal and two diagonal gives $2 \cdot 2^4 = 32$. Total $84$.

• Treating diagonally opposite sections as sharing a border, or forgetting that a section may repeat a neighbor's crop (only the two enemy pairs are forbidden).
• In Case 2, using $3$ options for $Q$ when the two fixed crops are enemies of each other; the forbidden crops are then $P$'s and $R$'s own crops, still two of them.