Stand Up & Math · Teacher Answers
The Starting Five
Not for distribution · Selections and arrangements · Full worked solutions
After two activities ran short, Round Table Trouble changed the recipe: the first idea in each phase is easy to spot, and the real work is justifying it for every case. This activity keeps that design. Dividing by \(r!\), choosing who sits out, subtracting unwanted cases and counting a pair two ways are all quick to find. What takes time is proving each one for general \(n\), the ratio argument in Phase 2, and the two-way counting in Phase 4. Do not accept “we tested it on a few rosters” as a proof. Hint ladders are for you only: give one rung at a time, and only if a group is stuck on the idea rather than on how to write it up. The challenge in Phase 4 is Pascal’s rule, which is proved again from lattice paths in Gridlock, so groups that reach it will see the same identity from two different directions.
“You’ve just been hired as head coach of a basketball club. Every week the league office needs to know how many different team sheets you could possibly submit, because their software has to reserve space for every one of them. There are different forms with different rules, and the office has been burned by formulas that only work for small rosters. This time every formula you give them must be justified for any roster size, not just checked on a few.”
Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.
| 0–2 min | Oral launch; random groups at whiteboards |
| 2–14 min | Phase 1, Lineup or Squad (1e is quick; the proof in 1d is where time goes) |
| 14–28 min | Phase 2, The Mirror (2d is the longest single task of the lesson) |
| 28–42 min | Phase 3, House Rules |
| 42–54 min | Phase 4, Captain’s Count |
| 54–60 min | Consolidation; challenge groups present |
Does the order matter?
Listing systematically (alphabetical order stops duplicates such as ABC and BAC from sneaking in):
Groups must physically list the 10 squads rather than jump to a formula. Watch for a group that lists ABC and BAC as different squads: that is the order-matters thinking from Task 1a leaking in, and it is exactly the confusion the phase is designed to surface.
So the lineup count is the squad count multiplied by \(r!\).
Every lineup sheet is made of two independent decisions: which \(r\) players are used (a squad), and in what order they fill the positions. So the lineup sheets split into groups, one group for each squad.
Every group has the same size: the \(r\) chosen players can be put in order in \(r!\) ways, whichever \(r\) players they are. So \({}^{n}P_{r} = {}^{n}C_{r} \times r!\), and dividing by \(r!\) gives the formula.
In Round Table the \(n!\) line-ups split into groups of size \(n\) (rotations) and dividing gave \((n-1)!\). Here the \({}^{n}P_{r}\) lineups split into groups of size \(r!\) (reorderings) and dividing gives \({}^{n}C_{r}\). Groups that name this link are showing real transfer. The key phrase to listen for is “every group has the same size”, because dividing is only legitimate when that is true.
Dividing by \(r\) instead of \(r!\). Have the group test it against their own listing: \(60 \div 3 = 20\), but they counted 10 squads in 1b. Choosing 4 from 6 fails the same way: \(360 \div 4 = 90\), not 15.
There is exactly one way to choose nobody (the empty squad sheet) and exactly one way to choose everybody, so \(\binom{n}{0} = \binom{n}{n} = 1\).
So the formula forces the convention \(0! = 1\); it is not an arbitrary rule.
Require: 60 and 10 obtained by listing, the relationship \({}^{n}P_{r} = {}^{n}C_{r} \times r!\) with the “each squad is counted \(r!\) times” reasoning in 1d, and 1e done with visible cancellation (no expanding 12!).
Who plays, and who sits
The row is symmetric, and the total is \(1+8+28+56+70+56+28+8+1\).
Do not name the “256 = 2⁸” link for them. Groups that spot it now are early for Phase 4b; ask them to hold the thought. The symmetry is what Task 2b is about.
In words: choosing \(r\) players to play is the same decision as choosing the \(n-r\) players who sit out. Every \(r\)-squad has exactly one complement, which is an \((n-r)\)-squad, and every \((n-r)\)-squad is the complement of exactly one \(r\)-squad. The two kinds of squad are matched one-to-one, so there are equally many.
Algebra:
Both proofs must be present. A common weak version of the word proof is “the row is symmetric” with no matching argument. Push for the one-to-one pairing, since that is the real reason.
The ratio exceeds 1 exactly when \(n - r > r + 1\), that is \(r < \dfrac{n-1}{2}\). So the row rises while \(r < \frac{n-1}{2}\), is level for one step if \(r = \frac{n-1}{2}\) (this only happens when \(n\) is odd, giving two equal middle terms), and falls after that.
Why it never rises again: \(\dfrac{n-r}{r+1}\) gets strictly smaller as \(r\) increases (the numerator falls and the denominator grows). Once it is 1 or less, every later ratio is below 1, so every later step is a fall.
This is the longest task of the lesson. Most groups will find the ratio, but many will stop at “it peaks in the middle”. The proof obligation is the never rises again part: it needs the observation that the ratio itself is decreasing. If a group has the peak but not that argument, ask “how do you know the row can’t go up again later?”.
Cancelling incorrectly to get \(\dfrac{n}{r+1}\) or \(\dfrac{n-r}{r}\). Have the group test their ratio on the row of 8: the ratio from 28 to 56 is 2, and \(\frac{8-2}{2+1} = 2\).
By 2d the row strictly increases up to the middle, may be level for one step, then strictly decreases. So any value occurs at most twice, at two positions on opposite sides of the middle, and those two positions are \(a\) and \(n-a\). Therefore \({}^{n}C_{a} = {}^{n}C_{b}\) with \(a \ne b\) forces \(a + b = n\): the Mirror rule is the only way.
Require: both proofs in 2b, 1225 / 4950 / \(n=15\) by hand, the simplified ratio \(\frac{n-r}{r+1}\), the correct peak, and an explicit “the ratio is decreasing” argument. A group can have the code before 2e if time is very short, but 2d must be solid.
Restrictions, and the order of the choosing
Every squad of 5 either includes S or excludes S, never both, so the two counts add to the total. The two counts are equal because \(\binom{9}{4} = \binom{9}{5}\), which is the Mirror rule from Phase 2.
This is Pascal’s rule for \(n = 10,\ r = 5\) in disguise: \(\binom{10}{5} = \binom{9}{4} + \binom{9}{5}\). Do not name it. Phase 4’s challenge asks students to prove it in general, and Gridlock proves it again from lattice paths.
Method 1, direct (cases by number of guards):
Method 2, complement:
“Choose 2 guards first (\(\binom{5}{2} = 10\)), then any 3 of the other 10 players (\(\binom{10}{3} = 120\)): \(10 \times 120 = 1200\).” This counts a squad with 3 guards three times (once for each pair of guards that could be “the first two”). It is the single most common error here, and it is worth having the group explain why 1200 is too big (it is bigger than 792, the total number of squads).
Both count the same set of outcomes (a squad together with a captain and a vice-captain from it). Each outcome is produced exactly once by either description, so the totals must agree.
This is the first time students see choose-then-arrange. Some groups will want to use \({}^{11}P_{2}\) in (i) as well; they are right that the roles are an ordered pick (captain and vice-captain are different jobs), and \(5 \times 4 = {}^{5}P_{2}\). I chose 11 players rather than 10 deliberately: with 10 the answer coincides with \({}^{10}P_{4} = 5040\) (the Vault Code number) by pure accident, which would hide wrong methods.
The product \(\binom{10}{2}\binom{10}{3}\) chooses the 3 others from all 10 people, so one of the captains can be picked again as one of the three. Those outcomes are not allowed (nobody is both). Choosing the 3 others from the 8 who are not captains fixes this:
The final expression treats \(a\) and \(b\) identically, so swapping them changes nothing. In the story: choosing the captains first and then the others, or the others first and then the captains, produces the same outcomes. The leftover \(n-a-b\) people form a third group (the ones who sit).
This expression is the number of ways to split \(n\) people into three labelled groups of sizes \(a,\ b,\ n-a-b\). In Gridlock, students meet the same numbers as the count of arrangements of a word with repeated letters. Do not mention that; it pays off next lesson.
Require: 126 / 126 / 252 with the in-or-out explanation, 596 by two methods, 9240 two ways, the overlap error identified in 3d (2520), and the factorial form in 3e. 3e can be left unfinished only if the class is running very late.
Counting the same thing twice
Team first: \(\binom{n}{k}\) ways to choose the team, then \(k\) ways to choose its captain: \(k\binom{n}{k}\) outcomes.
Captain first: \(n\) ways to choose the captain, then \(k-1\) teammates from the other \(n-1\) players: \(n\binom{n-1}{k-1}\) outcomes.
Both count exactly the same set of (team, captain) outcomes, so they are equal.
The key sentence is “we counted the same thing two ways”. A group that checks the equation numerically but never says why the two counts must agree has not proved anything. This technique, double counting, is the most powerful proof idea in the whole activity.
By size: \(\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n}\).
Player by player: each player is either in or out, 2 choices each, independently: \(2 \times 2 \times \cdots \times 2 = 2^n\) (the fundamental counting principle from the Vault Code).
For \(n = 8\) this is the 256 found in Task 2a.
Require: both double-counting arguments stated in words and the 256 link made explicitly.
(i) Squads of \(r\) that include the star S: choose the other \(r-1\) from the remaining \(n-1\), giving \(\binom{n-1}{r-1}\). Squads that exclude S: choose all \(r\) from the remaining \(n-1\), giving \(\binom{n-1}{r}\). Every squad is one or the other:
With \(n=10,\ r=5\) this is exactly Task 3a: \(252 = 126 + 126\).
(ii) Fix one particular player P. Toggling P (adding P if absent, removing P if present) turns every squad into a different squad whose size differs by exactly 1, so it changes even to odd and odd to even. Toggling twice returns the original, so this pairs up all \(2^n\) squads into even-odd pairs. Hence exactly \(2^{n-1}\) squads are even-sized and \(2^{n-1}\) are odd-sized.
Very few groups reach (ii) and that is fine; it is the release valve for groups that finish early. Part (i) is the main goal: it is Pascal’s rule, and it will be proved a second time from a completely different direction in Gridlock. If a group finishes both, ask them for the algebraic proof of Pascal’s rule using factorials and compare the two proofs.
1. Have a group present their Phase 1 proof. The phrase to draw out is “every squad is counted \(r!\) times, and all the same number of times”, and connect it to Round Table Trouble (dividing out a symmetry).
2. Ask the class: “How do you know the row of 8 can’t go up again after it starts falling?” The decreasing ratio is easy to state sloppily.
3. For 3b, ask a group to show both methods and then ask why 1200 is impossible (it is bigger than the total number of squads).
4. Have a group present 4a. Ask: “What does it mean to prove an equation by counting the same thing two ways?”
5. Tease Gridlock: “Next lesson a courier crosses a city, and these same numbers will show up somewhere you do not expect.”