Stand Up & Math · Teacher Answers
Round Table Trouble
Not for distribution · Circular permutations · Full worked solutions
After Vault Code ran short, the fix here isn't more separate tricks — it's tasks that stay hard after the trick is spotted. Three of the four phases explicitly require a written proof or a general formula, not just a number. Expect this to run long in a good way: proof-writing and generalization are slower, more effortful work than pattern-spotting, even for a group that finds the underlying idea quickly. Don't accept "we checked it for a few values and it worked" as a proof — push for a real argument, using the hint ladders below only if a group is stuck on the idea itself, not on how to write it up.
"You're planning the seating for a major round-table event. The client keeps adding rules, and every miscount means the wrong number of place settings gets ordered. This time, a number alone won't be enough — you'll need to convince the client your formula is always right, not just right for the case you happened to check."
Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.
Rethinking what "different" means
Wrapping each into a circle and reading "who's to the right":
So 6 linear arrangements collapse into just 2 distinct circular seatings — matching \((3-1)! = 2\).
Groups need to physically do this grouping exercise, not just guess the answer is "6 divided by something." Watching them notice that exactly 3 of the 6 line-ups collapse together (not 2, not 6) is the actual content of this phase.
Since rotating the entire table doesn't create a new arrangement, we can always describe a circular seating by first noting who sits in one fixed reference seat — say, imagine we always start our description from wherever a specific person, A, happens to be sitting. Fixing A's seat eliminates the redundant rotations from the count entirely: there is no way to "rotate into" a different description once A's position is nailed down.
With A's seat fixed, the remaining \(n-1\) people can be arranged in the remaining \(n-1\) seats in \((n-1)!\) ways, and every one of these gives a genuinely different circular seating (since A's position relative to every other person is now fixed and can't be rotated away). So there are exactly \((n-1)!\) distinct circular arrangements.
Every circular arrangement corresponds to exactly \(n\) different linear arrangements (one for each choice of where to "cut" the circle into a line). Since there are \(n!\) linear arrangements total, and each circular arrangement is counted \(n\) times among them, the number of distinct circular arrangements is \(n!/n = (n-1)!\).
Require: correct grouping in Task 1b (physically shown, not asserted), correct formula \((n-1)!\), and a genuine proof attempt in 1d — reject "it worked for n=3, 4, and 5 so it's probably always true" as insufficient; push for the general argument.
When flipping doesn't make it new
Flipping the table reverses the direction of every arrangement (clockwise becomes counterclockwise), but doesn't change who's sitting next to whom. Every circular arrangement therefore has exactly one distinct "mirror partner" — and since all \(n\) guests are distinct people, no arrangement can ever equal its own mirror image (that would require the seating to read identically forwards and backwards, which is impossible for \(n\ge3\) distinct individuals). So the \((n-1)!\) arrangements from Phase 1 pair up perfectly into mirror-pairs, and dividing by 2 removes exactly the duplicate.
Require: the pairing argument stated explicitly (not just "divide by 2 because mirror images"), and specifically the observation that arrangements pair up with NO leftover unpaired ones — this is what justifies dividing by exactly 2 rather than some other number.
When guests refuse to be separated
Treat A and B as one combined unit: now there are \(6-2+1=5\) units in the circle.
With \(k\) specific guests glued into one block, there are \(n-k+1\) total units in the circle (the block counts as 1 unit, plus the \(n-k\) individual remaining guests).
Verified against direct enumeration for several \((n,k)\) pairs, including \((6,2)\to48\), \((6,3)\to36\), \((7,3)\to144\) — all match.
Require: correct 48 for the specific case, and the general formula \(k!(n-k)!\) with the "units in the circle" reasoning stated explicitly — a group that jumps straight to the formula without explaining where \(n-k+1\) and \(k!\) come from hasn't actually generalized, just pattern-matched.
Two conditions at once
Step 1 — total with A&B together (Phase 3 method, \(n=8,k=2\)): \(2 \times 6! = 1440\).
Step 2 — subtract those where C&D are ALSO together: treat A&B as one block AND C&D as a second block, giving \(8-4+2=6\) total units.
Step 3 — subtract: \(1440 - 480\)
Verified directly against brute-force enumeration for \(n=8\): 960, exact match.
This is the real synthesis of the activity: block method (Phase 3) supplies the starting total, and complementary counting (the "at least/none" idea from the Vault Code activity) supplies the subtraction. A group that names both techniques explicitly in Task 4b has demonstrated real transfer across activities, not just within this one.
Require: correct starting total (1440), correct "both together" subtraction (480), correct final answer (960), and Task 4b explicitly naming the block method and complementary counting as the two ideas combined.
This requires genuine inclusion–exclusion, not just a second subtraction: subtracting "C&D together" and "E&F together" separately double-subtracts the case where BOTH extra pairs are together, so it must be added back once.
Very few groups will reach a fully correct inclusion–exclusion setup unprompted — that's expected and fine, it's the release valve for groups that finish everything else early. If a group subtracts both "bad" cases but forgets to add back the overlap, their answer will be too small; that's a good diagnostic moment to ask "have you removed anything twice?"
1. Have a group present their Phase 1 proof on the board — specifically the fix-one-seat argument — since it's the foundation everything else builds on.
2. Ask the class directly: "why does dividing by 2 in Phase 2 work perfectly, with nothing left over?" — the pairing argument is easy to state sloppily and worth making precise together.
3. For Phase 4, ask a group to explicitly name the two earlier phases they had to combine, and why neither alone was enough.
4. If any group reached the inclusion–exclusion challenge, have them present the "subtract twice, add back once" structure — this is the same idea they'll formalize properly if they continue into more advanced counting later.