2024 AMC 10B Problem 20

Record each position as $L$ or $R$. Whenever an $L$ and an $R$ are adjacent they must be mates. Since a shoe has only one mate, no shoe can be adjacent to two shoes of the opposite type: the patterns $LRL$ and $RLR$ cannot appear anywhere. Equivalently, every run of identical letters that is not at an end of the row has length at least $2$.

With three $L$'s and three $R$'s, the allowed patterns are:
- one boundary: $LLLRRR$, $RRRLLL$;
- two boundaries: $LLRRRL$, $LRRRLL$, $RRLLLR$, $RLLLRR$, $LRRLLR$, $RLLRRL$.

Now assign actual shoes. Each $L$–$R$ boundary forces the two shoes on either side to be a pair.

• One boundary (2 patterns): pick the pair straddling the boundary ($3$ ways). The other two left shoes fill the remaining $L$ slots in $2$ ways, the other two right shoes fill the remaining $R$ slots in $2$ ways: $3 \cdot 2 \cdot 2 = 12$ per pattern.
• Two boundaries (6 patterns): the two boundaries use two different pairs ($3 \cdot 2 = 6$ ways to assign them), and the third pair's shoes fill the two remaining slots, one $L$ and one $R$, in exactly $1$ way: $6$ per pattern.

Total: $2 \cdot 12 + 6 \cdot 6 = 24 + 36 = 60$.

The answer is $\boxed{\textbf{(A)}\ 60}$.

An adjacency restriction on labeled objects is easier to handle in two layers: first the coarse type pattern (here $L$/$R$), which the restriction almost determines, then the fine assignment, where the restriction becomes a set of forced pairings. Turning "no $L$ next to a foreign $R$" into "no $LRL$ or $RLR$" is the key reformulation.

Fix the first shoe ($6$ choices, all equivalent by symmetry) and enumerate the rest. If the second shoe is its mate, the third must be one of the $2$ remaining shoes of the second's type, and each choice finishes in $2$ ways: $4$ completions. If the second shoe is of the same type as the first ($2$ choices), each finishes in $3$ ways: $6$ completions. Total $6 \cdot (4 + 6) = 60$.

• Counting the L/R patterns but then assigning the six shoes freely (3! * 3!) instead of noticing that each boundary forces a specific pair to occupy both boundary slots.
• Reading the condition as "each shoe must be next to its mate," which is a different (and stricter) problem.