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Module 2 · Probability Foundations · PR-3
The reader said yes: he'll keep a two-week sleep diary. Maya sketches the follow-up piece and hits a wall immediately — one week of nights, each either short or okay, can unfold in more ways than she can list, and she still has to schedule interviews and pick a feature group from her volunteers. Earlier in this module, listing outcomes worked. Today it stops working — and three formulas take over.
Live session · join with code OTTO-PR3 · ottolearn.org/join
Same rules as always: vote before I explain, every poll is anonymous, and changing your mind after discussion is the goal.
Today · the diary follow-up Maya is planning
The Fundamental Counting Principle
Factorials — the arrange-everything number
Permutations — when order is part of the outcome
Combinations — when only membership matters
The swap test, and probability from counts
Five questions, three formulas, one decision test. Every one starts with your vote.
Recall · last class · 45 seconds
Warm-up · 60 seconds · anonymous
Hold your vote — the first concept settles it. And notice the stakes: if Maya can't count the possible weeks, she can't say whether the reader's week was unusual.
Core concept 1 · C1
Maya's diary week is a process in stages: night 1 has 2 outcomes; for each of those, night 2 has 2; for each of those pairs, night 3 has 2 — and so on.
A multi-stage process with \(m_1\) choices at stage 1, then \(m_2\), then \(m_3\)… has \(m_1 \times m_2 \times \cdots\) complete sequences. The diary week: \(2 \times 2 \times \cdots \times 2 = 2^7 = 128\) patterns — the warm-up, settled.
\(m_1 \times m_2 \times \cdots \times m_k\)Each stage-1 choice pairs with every stage-2 choice — a tree fanning out. Adding counts the branches at one stage; multiplying counts the complete paths, and an outcome is a complete path.
The count at each stage must not depend on earlier choices. A 4-digit code with repeats allowed: \(10^4\). Require four different digits and the stages start shrinking — \(10 \times 9 \times 8 \times 7\). That shrinking pattern gets its own name in concept 3.
One habit: read any counting problem as stages, and ask what each stage offers. If the offers never shrink, the counting principle finishes the job on its own.
See it · C1
Predict first: the tree shows every complete path. Set stage 1 to 4 and stage 2 to 3 — how many paths appear? Then switch on No repeats — what happens to each stage's count, and why?
Interactive tree diagram for the Fundamental Counting Principle. Use the buttons to set the number of choices at Stage 1 and Stage 2, optionally add a Stage 3, and switch on no-repeats mode to make each stage draw from a depleting pool. The tree, its description, and the running product update automatically.
Same pool, no item reused — each stage has one fewer choice.
Practice · the counting principle · fresh numbers on demand
Core concept 2 · C2
Line up all \(n\) objects: \(n\) choices for first place, \(n-1\) for second, down to 1. It's the counting principle with shrinking stages — and it earns its own symbol because it appears everywhere.
\(n! = n \times (n-1) \times \cdots \times 1\)\(3! = 6\). \(5! = 120\). \(10! = 3{,}628{,}800\) — ten people can line up in more ways than Maya's campus has students, times 400. This explosion is why listing fails and formulas take over.
There is exactly one way to arrange nothing: do nothing. The two formulas ahead divide by factorials, and \(0! = 1\) is what keeps them from breaking at the edges — it is never 0.
\(0! = 1\)Factorials are rarely the final answer — they're the gears inside the two formulas coming next. Watch how they cancel:
See it · C2
Predict first: expand \(\frac{n!}{(n-r)!}\) into its terms, then click Show cancellation. How many terms survive the strike-through? Count before you click.
Interactive fraction showing n! divided by (n minus r) factorial, expanded as individual terms. Choose n and r, then click "Show cancellation" to see the denominator terms strike through against the matching numerator terms, leaving the r highlighted terms that multiply to give the permutation count.
Core concept 3 · C3
The plan grows: 10 volunteers from the survey signed up for diary follow-ups. Maya must fill 3 interview slots — Monday, Tuesday, Wednesday.
Monday's slot: 10 candidates. Tuesday: 9 remain. Wednesday: 8. The counting principle with the shrinkage built in — each choice uses someone up.
\(10 \times 9 \times 8 = 720\)Selecting \(r\) of \(n\) objects in order: start from \(n!\) and cancel the unused tail, \((n-r)!\). The cancellation you just watched is this formula at work.
\(_nP_r = \dfrac{n!}{(n-r)!}\)Only when swapping two selections gives a different outcome. Ali on Monday, Ben on Tuesday is a different schedule from Ben on Monday, Ali on Tuesday — order is part of the outcome here.
Signal words: arrange, rank, schedule, sequence, 1st–2nd–3rd. But signal words can lie — the reliable test comes at concept 5.
See it · C3
Predict first: each position shows a live choices badge. Fill the slots one click at a time — what number will the badge show at the third position, and where does that number come from?
Interactive visualization with 5 labelled objects A through E. Click an object to place it in the next position slot. The badge above each slot shows how many choices were available at that step. Click a filled slot to remove it and all slots after it. Use the r tabs to change the number of positions.
We do · you finish it
Maya's editor asks: "How many different schedules for the three interview days can you build from the 10 volunteers?"
Finish it: \(_{10}P_3 = ?\)
Practice · permutations · fresh numbers on demand
Core concept 4 · C4
Plan B: no interview days. Maya just needs a group of 3 volunteers to feature in the piece — no slots, no order, no roles.
The 720 schedules keep re-listing the same trio: Ali–Ben–Cléo shows up in \(3! = 6\) different orders, but it's one group. Every trio is counted exactly 6 times.
Count ordered, then divide by the orderings each group has: \(720 \div 6 = 120\) groups. That division is the combination formula — \(_nC_r = {}_nP_r / r!\), always.
\(_nC_r = \dfrac{n!}{r!\,(n-r)!}\)Choosing 3 to feature is the same act as choosing 7 to leave out: \(\binom{10}{3} = \binom{10}{7} = 120\). When \(r\) is large, count the small side instead.
\(\dbinom{n}{r} = \dbinom{n}{n-r}\)Permutation and combination aren't rivals — a combination is a permutation with the order divided out. The only question is whether order deserved to be counted.
See it · C4
Predict first: the left panel lists ordered arrangements, tinted by which objects they contain. Switch to Grouped — how many orderings fall into each group, and what does that number have to do with \(r!\)?
Interactive comparison for 4 objects A, B, C, D at r = 2. The left panel lists all 12 ordered arrangements (permutations); the right panel lists all 6 unordered selections (combinations). Every group of 2! = 2 orderings of the same objects collapses to one combination: 12 divided by 2! equals 6. Use the r tabs to change the selection size and the group view control to tint or collapse the groups.
We do · you finish it
Same 10 volunteers — but now Maya only needs a group of 3, no interview days attached.
Finish it: \(\dbinom{10}{3} = ?\)
Commit first · today's summit
Core concept 5 · C5
Swap two of the selected items and ask: is the outcome different? Different → order matters → permutation. Same → only membership matters → combination.
Schedules, rankings, codes, playlists — being first is different from being second. The keypad: 1-5-8-2 opens and 2-5-8-1 fails, so the "combination" lock counts permutations.
\(_nP_r\)Teams, committees, samples, toppings — the same members in any listing order are one outcome. Maya's feature group of 3 is a combination, whatever order she writes the names in.
\(_nC_r\)"Combination lock" proves the vocabulary can't be trusted. Run the swap test every time — it never lies.
I do · watch the whole decision
The editor asks for (a) a running order for 4 of Maya's 9 story ideas in the next issue, and (b) 3 of her 10 volunteers to receive thank-you gift cards. Before any formula, I say what one outcome is: (a) an ordered lineup of 4 ideas; (b) a set of 3 names.
(a) Swap ideas #1 and #4 in the lineup — the issue reads differently: different outcome, order matters. (b) Swap two names on the gift list — the same three people get cards: same outcome, order is noise.
(a) Permutation: \(_9P_4 = 9 \times 8 \times 7 \times 6 = 3024\) running orders. (b) Combination: \(\binom{10}{3} = \dfrac{10 \times 9 \times 8}{3!} = 120\) gift lists.
Ordered counts always dwarf their unordered twins — bigger by exactly \(r!\). Check: \(\binom{9}{4} = 126\) and \(126 \times 4! = 3024\) ✓. If a "permutation" ever comes out smaller than the matching combination, a formula got swapped.
The habit: outcome first, swap test second, formula last. Never let a scenario's vocabulary pick the formula for you.
See it · C5
Trace the keypad code and the gift list through the chart before expanding the worked checks — do both land where the swap test said they would?
Swapping president and VP gives a different outcome — order matters, so
Permutation.
12P3 = 12 × 11 × 10 = 1320
{Ann, Bob, Carl} and {Carl, Ann, Bob} are the same committee —
order doesn't matter, so Combination.
C(12, 3) = 12! / (3! · 9!) = 220
Note: 220 = 1320 ÷ 3! — the same 12 · 11 · 10 selections, with each group's 6 orderings collapsed
into one.
Practice · which tool? · fresh numbers on demand
Core concept 6 · C6
Equally-likely outcomes: favorable count over total count — the first formula of this module. Nothing about it changes; only the counting gets industrial.
\(P(A) = \dfrac{n(A)}{n(S)}\)If the denominator counts unordered groups, the numerator must count unordered groups too. Mixing a permutation over a combination silently inflates the answer by \(r!\) — the flaw hunt in a moment turns on exactly this.
Inside a favorable count, independent choices multiply: picking 2 part-timers and 1 other volunteer is \(\binom{4}{2} \times \binom{6}{1}\) — the counting principle working inside a combination count.
Maya's group of 3 from her 10 volunteers: all \(\binom{10}{3} = 120\) groups equally likely. Which means she can now put a probability on who ends up featured.
See it · C6
The same move at lesson scale: 3 students drawn from a class of 20, split by how many of the 8 athletes are included. Predict first: which segment is bigger — "exactly 1 athlete" or "no athletes"? Click the segments to check the formula behind each.
Proportional bar showing all four outcomes when choosing 3 students from 20 (8 athletes, 12 non-athletes). Each segment width is proportional to the number of equally-likely committees in that category. Click a segment to see the formula and probability.
Click a segment to see the probability formula
Find the flaw · one line is wrong
Which tool decides? · mixes earlier classes + today
See it · C6
The bar splits all committees by how many athletes they include — the lesson's bigger example again. Predict first: which single slice is the complement of "at least one athlete"? Everything else needs no counting at all.
Proportional bar dividing all 1,140 equally-likely committees of 3 students (from 20: 8 athletes, 12 non-athletes) into two segments. The small segment, 0 athletes, has C(12,3) = 220 committees, about 19.3 percent. The large segment, at least one athlete, has 1,140 minus 220 = 920 committees, about 80.7 percent. So P(at least one athlete) = 1 minus 220 over 1140 = 920 over 1140, approximately 0.807. Faint dividers inside the large segment mark the exactly-1, exactly-2, and exactly-3 athlete cases (528, 336, and 56 committees) that the complement strategy lets you skip: counting the one small none slice is easier than summing three separate cases.
P(at least one) = 1 − 220/1140 = 920/1140 ≈ 0.807
Counting the one small "none" slice — C(12,3) = 220 — takes a single combination. The faint lines inside the big segment mark the exactly-1 / exactly-2 / exactly-3 cases (528, 336, 56) you did not have to compute.
Practice · counting probabilities · fresh numbers on demand
Production task · write the reply
Exit ticket · 60 seconds · anonymous
What Maya files
Stages multiply: a diary week has \(2^7 = 128\) possible patterns — counted in seconds, listed never.
One test picks the formula: swap two selections. Different outcome → \(_{10}P_3 = 720\) schedules; same outcome → \(\binom{10}{3} = 120\) groups. Vocabulary lies; the swap test doesn't.
Counts become probabilities: same method top and bottom, and "at least one" flips to one subtraction — \(P \approx 0.83\) that a part-timer's voice makes the piece.
The reader's week: 35 of the 128 patterns — if every pattern is equally likely. That "if" is a claim about how random nights actually behave. Next class: random variables and probability distributions — putting honest weights on the 128.
Vote counts on today's slides are simulated for rehearsal; live sessions show the room's real votes.