2021 AMC Fall 10A Problem 18

Each crop has exactly one crop it cannot sit next to: corn and wheat are "enemies", and so are soybeans and potatoes. Any crop can sit next to a copy of itself.

Name the sections: top-left $P$, top-right $Q$, bottom-right $R$, bottom-left $S$. Then $P$ and $R$ are diagonal and do not share a border, and neither do $Q$ and $S$. But $Q$ touches both $P$ and $R$, and so does $S$.

So choose the crops for $P$ and $R$ first. Then $Q$ and $S$ each only need to get along with $P$ and $R$, and do not affect each other.

Case 1: $P$ and $R$ get the same crop. There are $4$ ways (for example, both corn). Now $Q$ must only avoid that crop's enemy (wheat), leaving $3$ choices. $S$ also has $3$ choices. Total: $4 \times 3 \times 3 = 36$.

Case 2: $P$ and $R$ get different crops. There are $4 \times 3 = 12$ ways. Now $Q$ must avoid two enemies, one for each of them. For example, if $P$ is corn and $R$ is soybeans, $Q$ cannot be wheat or potatoes. Different crops always have different enemies, so exactly $2$ crops are ruled out and $Q$ has $2$ choices. $S$ also has $2$. Total: $12 \times 2 \times 2 = 48$.

Add the cases. $36 + 48 = 84$.

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

In a $2 \times 2$ grid, the two diagonal sections are the ones that do not touch. Once you fix them, the other two sections face the same rule and do not bother each other, so you can just multiply. Splitting into "diagonal pair same" and "diagonal pair different" keeps it to two short cases. In grid coloring problems, look for cells that do not touch and fill them first.

Sort the crops into two families: grains (corn, wheat) and roots (soybeans, potatoes). Two neighbors from different families are always fine. Two neighbors from the same family must be the same crop. So first decide each section's family, then pick crops:
- All four in one family: they must all be the same crop. $2$ families $\times$ $2$ crops $= 4$.
- Three in one family, one in the other: $4$ choices for the odd section and $2$ for which family it is, so $8$ patterns. The three connected sections share one crop ($2$ ways), and the odd one has $2$ ways. $8 \times 2 \times 2 = 32$.
- Two side-by-side sections in each family: $4$ patterns, each block has $2$ ways. $4 \times 2 \times 2 = 16$.
- Families on the diagonals: $2$ patterns, and no two same-family sections touch, so each section has $2$ ways. $2 \times 2^4 = 32$.

Total: $4 + 32 + 16 + 32 = 84$.

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 $\boxed{\textbf{(C)}\ 84}$.