AMC 10 · 2024 · #20

• Reframe the rule on the L/R type-string.
• Whenever the type-string has an $L$ directly followed by an $R$ (or vice versa) — a "boundary" — those two shoes must be the matched $L_i, R_i$ for some common $i$.
• Key consequence: a type-substring $LRL$ (or $RLR$) is impossible, because the middle shoe would have to be both $R_i$ and $R_j$ for two different boundary-pair partners — same shoe in two positions.

💡 Strip the names off the shoes and look only at the L/R skeleton — the forbidden adjacency becomes a clean rule on a short string.

• List every type-string of three $L$'s and three $R$'s with no $LRL$ or $RLR$ substring.
• The allowed strings are exactly those whose consecutive-block shape is $\{3\}{\{3\}}$, $\{a, b, c\}$ with three blocks in $L,R,L$ or $R,L,R$ order, or $\{1, 2, 2, 1\}$ in $L,R,L,R$ or $R,L,R,L$ order.
• Enumerating by block shape: $LLLRRR, RRRLLL$ (2 strings, $1$ boundary each); $LLRRRL, LRRRLL, RLLLRR, RRLLLR$ (4 strings, $2$ boundaries each); $LRRLLR, RLLRRL$ (2 strings, $3$ boundaries each).
• $8$ valid type-strings in all.

💡 Listing block-shapes systematically prevents both missing strings and double-counting; the four shapes are the only ones consistent with the $LRL/RLR$ ban.

• Subproblem A — type-strings with $1$ boundary ($LLLRRR, RRRLLL$).
• For $LLLRRR$: the single boundary at positions $3, 4$ must be a matched pair $L_a R_a$, chosen in $3$ ways.
• The remaining two $L$'s go in positions $1, 2$ in $2! = 2$ orders; similarly for the right shoes in positions $5, 6$.
• Each shape contributes $3 \cdot 2 \cdot 2 = 12$.
• By symmetry $RRRLLL$ also gives $12$.
• Subtotal $24$.

💡 Once you fix which matched pair sits at the boundary, the rest is two independent small orderings — Tool #9's "reduce to smaller pieces" move.

• Subproblem B — type-strings with $2$ boundaries ($LLRRRL, LRRRLL, RLLLRR, RRLLLR$).
• Take $LLRRRL$: boundaries at $(2,3)$ and $(5,6)$ each force a matched pair, and the two pairs must be different (a shoe cannot appear twice).
• Pick the ordered pair of distinct pairs: $3 \cdot 2 = 6$ ways.
• Once chosen — say $(L_1, R_1)$ at $(2,3)$ and $(R_2, L_2)$ at $(5,6)$ — every position is forced: position $1$ is the remaining $L_3$ and position $4$ is the remaining $R_3$.
• Each of the four type-strings contributes $6$, subtotal $4 \cdot 6 = 24$.

💡 Two boundaries each "lock" one matched pair in place; the leftover positions for the third pair are then forced.

• Subproblem C — type-strings with $3$ boundaries ($LRRLLR, RLLRRL$).
• Take $LRRLLR$: boundaries at $(1,2), (3,4), (5,6)$ each force a matched pair, and the three matched pairs must be a permutation of $\{P_1, P_2, P_3\}$.
• There are $3! = 6$ orderings, and every internal same-type adjacency (e.g.
• $R_1 R_2$, $L_2 L_3$) is permitted.
• Each type-string contributes $6$, subtotal $2 \cdot 6 = 12$.

💡 Three boundaries means each boundary uses one pair, and the three pairs are just permuted — $3! = 6$.

Sum the three subtotals.

💡 Adding three disjoint subtotals is Grade 4 multi-digit addition — and the cases are guaranteed disjoint because each shoe ordering has one and only one L/R type-string.