AMC 10 · 2024 · #20
Grade 7 countingProblem
Three different pairs of shoes are placed in a row so that no left shoe is next to a
right shoe from a different pair. In how many ways can these six shoes be lined up?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three different pairs of shoes — call them $P_1, P_2, P_3$, each with a left shoe $L_i$ and a right shoe $R_i$ — are placed in a row of $6$ positions. The rule: no left shoe is next to a right shoe from a different pair (i.e., an $L_i R_j$ or $R_j L_i$ adjacency is allowed only if $i = j$). How many orderings of the six distinct shoes obey the rule?
Givens: $6$ distinct shoes: $L_1, R_1, L_2, R_2, L_3, R_3$; Forbidden: adjacent pair $L_i \;\&\; R_j$ with $i \ne j$; Allowed: any two $L$'s adjacent, any two $R$'s adjacent, or a matched $L_i R_i$ adjacency; Answer choices: (A) $60$, (B) $72$, (C) $90$, (D) $108$, (E) $120$
Unknowns: Number of valid orderings of the six shoes in a row
Understand
Restated: Three different pairs of shoes — call them $P_1, P_2, P_3$, each with a left shoe $L_i$ and a right shoe $R_i$ — are placed in a row of $6$ positions. The rule: no left shoe is next to a right shoe from a different pair (i.e., an $L_i R_j$ or $R_j L_i$ adjacency is allowed only if $i = j$). How many orderings of the six distinct shoes obey the rule?
Givens: $6$ distinct shoes: $L_1, R_1, L_2, R_2, L_3, R_3$; Forbidden: adjacent pair $L_i \;\&\; R_j$ with $i \ne j$; Allowed: any two $L$'s adjacent, any two $R$'s adjacent, or a matched $L_i R_i$ adjacency; Answer choices: (A) $60$, (B) $72$, (C) $90$, (D) $108$, (E) $120$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
The shoes look complicated, but the only thing the rule cares about is *boundaries between L's and R's* in the row. Tool #2 (Make a Systematic List) gives a clean two-layer plan: layer one — list every $L/R$ type-string of three $L$'s and three $R$'s that has no forbidden boundary; layer two — for each surviving type-string, count how many ways to label the $L$'s as $L_1, L_2, L_3$ and the $R$'s as $R_1, R_2, R_3$ so each $L$-$R$ boundary is a matched pair. Tool #7 (Identify Subproblems) splits the type-strings into four block-shape cases. Tool #9 (Solve an Easier Related Problem) reduces the labeling layer to a small permutation count per case.
Execute — Answer: A
5.OA.A.2 Step 1 - 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.
7.SP.C.8 Step 2 - 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.
4.OA.A.3 Step 3 - 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.
7.SP.C.8 Step 4 - 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.
4.OA.A.3 Step 5 - 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 — $3! = 6$.
4.NBT.B.4 Step 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.
5.OA.A.2 Reframe the rule on the L/R type-string. Whenever the type-string has an $L$ dir 7.SP.C.8 List every type-string of three $L$'s and three $R$'s with no $LRL$ or $RLR$ sub 4.OA.A.3 Subproblem A — type-strings with $1$ boundary ($LLLRRR, RRRLLL$). For $LLLRRR$: 7.SP.C.8 Subproblem B — type-strings with $2$ boundaries ($LLRRRL, LRRRLL, RLLLRR, RRLLLR 4.OA.A.3 Subproblem C — type-strings with $3$ boundaries ($LRRLLR, RLLRRL$). Take $LRRLLR 4.NBT.B.4 Sum the three subtotals. Review
Reasonableness: Cross-check the totals by an independent route. Each of the $8$ valid type-strings was placed in one of three classes — $1$, $2$, or $3$ boundaries — and the per-string counts $12, 6, 6$ depend only on the number of boundaries: $12 = 3 \cdot (2!)^2$ for one boundary, $6 = 3 \cdot 2$ for two boundaries (ordered pair of distinct pairs), $6 = 3!$ for three boundaries. Aggregating: $2 \cdot 12 + 4 \cdot 6 + 2 \cdot 6 = 24 + 24 + 12 = 60$. The total $60$ is well below the unconstrained $6! = 720$, consistent with a real adjacency restriction. Answer (A) $= 60$.
Alternative: Tool #16 (Change Focus / Complement). Treat the row as a sequence of moves between $L$- and $R$-blocks. Total arrangements with no constraint: $6! = 720$. Inclusion-exclusion on the forbidden boundaries (an $L_i$ adjacent to some $R_j$ with $j \ne i$, for each of the $5$ adjacent slots and the $6$ choices of mismatched pair) reproduces $60$. The casework above is faster but the complement view is a useful cross-check.
CCSS standards used (min grade 7)
5.OA.A.2Write simple expressions that record calculations with numbers (Reframing the rule on the $L/R$ type-string: at every $L$-$R$ boundary the two shoes must be a matched pair $L_i R_i$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Enumerating the $8$ valid $L/R$ type-strings of three $L$'s and three $R$'s with no $LRL$ or $RLR$ substring, and counting boundary-pair choices via $P(3, 2)$ in the two-boundary class.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Computing per-string counts $3 \cdot 2! \cdot 2! = 12$ for one-boundary strings and $3! = 6$ for three-boundary strings.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Summing the three subtotals $24 + 24 + 12 = 60$.)
⭐ This AMC 10 problem only needs Grade 7 organized counting — list every valid $L/R$ skeleton, then count name choices at each boundary — and the answer drops out as $60$!
⭐ This AMC 10 problem only needs Grade 7 organized counting — list every valid $L/R$ skeleton, then count name choices at each boundary — and the answer drops out as $60$!
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