AMC 10 · 2008 · #21
Grade 7 countingTen chairs are evenly spaced around a round table and numbered clockwise from 1 through 10. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?
Pick an answer.
AMC 10 2008 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: Ten chairs, numbered $1$ through $10$ clockwise, sit evenly spaced around a round table. Five married couples (five men and five women) take the chairs so that genders alternate around the circle. No person may sit in a chair next to their spouse, and no person may sit directly across the table from their spouse. Count how many seating arrangements obey all of these rules.
Givens: There are $10$ numbered chairs around a round table; because the chairs are numbered, two seatings that differ by a rotation are counted as different.; Five married couples supply five men and five women, all distinct people.; Seats alternate man, woman, man, woman, ... all the way around.; No one may sit in either chair immediately next to their spouse.; No one may sit directly across the table from their spouse (chairs $i$ and $i+5$ are 'across').
Unknowns: The number of valid seating arrangements of the ten people.
Understand
Restated: Ten chairs, numbered $1$ through $10$ clockwise, sit evenly spaced around a round table. Five married couples (five men and five women) take the chairs so that genders alternate around the circle. No person may sit in a chair next to their spouse, and no person may sit directly across the table from their spouse. Count how many seating arrangements obey all of these rules.
Givens: There are $10$ numbered chairs around a round table; because the chairs are numbered, two seatings that differ by a rotation are counted as different.; Five married couples supply five men and five women, all distinct people.; Seats alternate man, woman, man, woman, ... all the way around.; No one may sit in either chair immediately next to their spouse.; No one may sit directly across the table from their spouse (chairs $i$ and $i+5$ are 'across').
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #2 Make a Systematic List
The men's rule and the women's rule are tangled together, so tool #7 (Identify Subproblems) splits the count into two clean stages: first seat the five men using only the alternating rule, then, with the men frozen in place, count how many ways the five wives can fill the opposite-gender chairs without landing next to or across from their husbands. Tool #1 (Draw a Diagram) makes the geometry concrete: mark which chairs are neighbors and which are across, so each man's three forbidden chairs are easy to read off. Tool #2 (Make a Systematic List) then shows that every wife has only two legal chairs, and those choices lock together into a single loop, so only two full women-arrangements survive. Multiplying the men-count by the women-count gives the answer.
Execute — Answer: C
7.SP.C.8 Step 1 Seat the five men
- Because genders alternate, the men occupy either all five odd chairs ($1,3,5,7,9$) or all five even chairs ($2,4,6,8,10$) — that is $2$ choices for which chairs the men use.
- Within the chosen five chairs, the five distinct men can be ordered in $5! = 120$ ways.
- So the men alone can be seated in $2 \times 120 = 240$ ways.
💡 Pick the men's chairs first, then order the men in them; the two independent choices multiply.
7.SP.C.8 Step 2 Find each wife's legal chairs
- Freeze one case with a diagram: men in chairs $1,3,5,7,9$, women in $2,4,6,8,10$.
- The man in chair $1$ forbids his wife from chairs $2$ and $10$ (neighbors) and chair $6$ (across), leaving only chairs $4$ and $8$.
- Doing the same for every man leaves each wife exactly two legal chairs: man $1\to\{4,8\}$, man $3\to\{6,10\}$, man $5\to\{8,2\}$, man $7\to\{10,4\}$, man $9\to\{2,6\}$.
💡 Each man blocks three chairs for his wife, so out of the five opposite-gender chairs only two remain open for her.
7.SP.C.8 Step 3 Chain the wives: only two arrangements
- The legal chairs interlock, so one choice forces all the rest.
- Put man $1$'s wife in chair $4$: then chair $4$ is gone, so man $7$ (who could use $10$ or $4$) must take $10$; that takes $10$, so man $3$ takes $6$; then man $9$ takes $2$; then man $5$ takes $8$ — all consistent.
- Put man $1$'s wife in chair $8$ instead and the same forcing runs the other way: man $5\to 2$, man $9\to 6$, man $3\to 10$, man $7\to 4$.
- Those are the only two ways the wives can be placed.
💡 With two options each and every choice knocking out the next person's option, the whole ring settles into just two mirror-image fillings.
4.OA.A.3 Step 4 Multiply the two stages
- For every one of the $240$ ways to seat the men, there are exactly $2$ ways to seat the wives, and the two stages are independent, so multiply them: $240 \times 2 = 480$.
- That is choice $(C)$.
💡 Each men-seating pairs with each of the two women-fillings, so total arrangements is the product of the stage counts.
7.SP.C.8 Because genders alternate, the men occupy either all five odd chairs ($1,3,5,7,9 7.SP.C.8 Freeze one case with a diagram: men in chairs $1,3,5,7,9$, women in $2,4,6,8,10$ 7.SP.C.8 The legal chairs interlock, so one choice forces all the rest. Put man $1$'s wif 4.OA.A.3 For every one of the $240$ ways to seat the men, there are exactly $2$ ways to s Review
Reasonableness: Without the spouse rule, alternating seatings number $2 \times 5! \times 5! = 2 \times 120 \times 120 = 28800$; the 'no neighbor, no across' rule slashes each block of $120$ women-orderings down to just $2$, giving $240 \times 2 = 480$. The result is a positive whole number, it is a multiple of the men-count $240$ exactly as the '$\times 2$ for the women' structure predicts, and among the choices only $480$ is such a multiple of $240$ — all consistent with choice $(C)$.
Alternative: Seat the men one at a time instead of by parity. The first man may pick any of the $10$ chairs, which fixes the whole alternating pattern; the remaining four men fill the other same-gender chairs in $4 \times 3 \times 2 \times 1 = 24$ ways, again $10 \times 24 = 240$. The wives still admit exactly $2$ fillings, so $10 \times 4 \times 3 \times 2 \times 1 \times 2 = 480$, the same answer.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the men-seatings with the multiplication principle and using an organized list of each wife's legal chairs to show only two women-arrangements survive.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the two independent stage counts by multiplying $240 \times 2 = 480$.)
⭐ Split a tangled seating count into stages — place one group first, then count how the constraints squeeze the second group — and multiply the stage counts together.
⭐ Split a tangled seating count into stages — place one group first, then count how the constraints squeeze the second group — and multiply the stage counts together.
More like this
Same archetype — closest grade level first.