Combinatorics bundle + fix for daily-run duplication (per-subtopic numbering & coordination protocol)

CONTENT_SCHEMA.md

@@ -56,7 +56,7 @@ license: CC-BY-SA-4.0
 
 | Field | Required | Type | Notes |
 |-------|----------|------|-------|
-| `id` | yes | string | Unique. Must equal the filename without `.md`. Format: `<section>-<topic>-<subtopic>-NNNN`. |
+| `id` | yes | string | Unique. Must equal the filename without `.md`. Format: `<section>-<topic>-<subtopic>-NNNN`. `NNNN` is a 4-digit sequence number **unique within the subtopic** — pick the next unused number for that exact subtopic (a brand-new subtopic starts at `0001`). Per-subtopic numbering keeps parallel branches from colliding; see [CONTRIBUTING.md](CONTRIBUTING.md#file-naming--location). |
 | `section` | yes | enum | `quant` \| `verbal` \| `data-insights`. |
 | `topic` | yes | string | A topic slug from [`curriculum.md`](curriculum.md). |
 | `subtopic` | yes | string | A subtopic slug from [`curriculum.md`](curriculum.md). |

CONTRIBUTING.md

@@ -70,10 +70,19 @@ content/lessons/<section>/<topic>-<subtopic>.md
 
 - `<section>` is `quant`, `verbal`, or `data-insights`.
 - `<topic>` / `<subtopic>` must be slugs that exist in [`curriculum.md`](curriculum.md).
-- `NNNN` is a zero-padded sequence number, unique within the section (pick the next unused one).
+- `NNNN` is a zero-padded sequence number, **unique within the subtopic** — pick the next unused
+  number *for that exact subtopic*. A brand-new subtopic starts at `0001`; a subtopic that already
+  has `...-0001` and `...-0002` continues at `0003`.
 
 Example: `content/questions/quant/quant-algebra-linear-equations-0001.md`
 
+> **Why per-subtopic and not per-section?** Contributors (and automated runs) work on separate
+> branches that haven't merged yet. A single section-wide counter makes every open branch grab the
+> same "next" number, so two PRs collide the moment they touch the same section. Numbering *within
+> the subtopic* means two PRs only ever collide if they target the **same subtopic** — which the
+> coordination rules below already tell you to avoid. Existing files are already unique within their
+> subtopic, so nothing needs renaming.
+
 ---
 
 ## The review workflow
@@ -119,6 +128,33 @@ If unsure, tag `medium` and note your uncertainty in the PR — reviewers can ad
 
 ---
 
+## Automated & daily-run contributions
+
+Some content is added by automated "daily run" agents. Because every run works on its **own
+branch** and those branches are **not merged between runs**, a run that only looks at `main` is
+blind to what other in-flight runs have already produced — which historically led to several PRs
+re-seeding the same subtopic and colliding on sequence numbers. If you run one of these agents (or
+review their output), follow this protocol so parallel work doesn't duplicate:
+
+1. **Check what's already in flight, not just `main`.** Before choosing a subtopic, list the
+   repository's **open pull requests** and read their titles/branches. Treat any subtopic that an
+   open PR already covers as *taken* — do not seed it again. (`main` alone is not the current state
+   of the world; most content lives in unmerged branches.)
+2. **One subtopic per run, and prefer genuine gaps.** Pick a subtopic that is uncovered on `main`
+   **and** not claimed by any open PR. Run `node scripts/coverage.mjs` to see per-subtopic counts
+   and gaps on the current checkout, then cross-check against open PRs for the in-flight ones.
+3. **Number within the subtopic.** Follow the [file-naming rule](#file-naming--location):
+   `NNNN` is the next unused number *for that subtopic*. Since rule 2 keeps runs on distinct
+   subtopics, per-subtopic numbering makes filename collisions between branches impossible without
+   having to inspect other branches' numbers.
+4. **Keep each run on its own branch and its own PR.** Don't stack a new run's content onto another
+   run's branch, and don't reuse a merged PR — start fresh from the latest default branch.
+5. **Stop at "open PR."** Opening the draft PR is the end of a run's job. Don't re-seed a subtopic
+   that any open PR (including your own from a previous run) already handles.
+
+Reviewers/maintainers: merging or closing the open backlog promptly is what makes rules 1–2
+reliable — the longer PRs sit unmerged, the more the "check open PRs" step has to compensate for.
+
 ## Code of conduct
 
 Participation is governed by our [Code of Conduct](CODE_OF_CONDUCT.md). Be kind; assume good

content/lessons/quant/quant-counting-probability-combinatorics.md

@@ -0,0 +1,88 @@
+---
+id: quant-counting-probability-combinatorics
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+title: "Combinatorics: Counting Without Listing"
+tags: [combinatorics, counting, permutations, combinations, factorials]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Combinatorics is the art of counting arrangements and selections without writing them all out.
+Almost every Focus counting problem reduces to two questions: **do I multiply or add?** and **does
+order matter?** Get those two right and the formulas fall into place.
+
+## Core concepts
+
+**Fundamental counting principle (multiply for "and").** If one choice can be made \(m\) ways and
+an independent next choice \(n\) ways, the two together can be made \(m \times n\) ways. Multiply
+stages that happen *together*; **add** only mutually exclusive *cases* ("or").
+
+**Factorial — arranging distinct things in a row.** \(n\) distinct items in a row:
+
+\[n! = n \times (n-1) \times \cdots \times 2 \times 1\]
+
+By convention \(0! = 1\).
+
+**Permutations — order matters.** Choosing and arranging \(k\) of \(n\) distinct items:
+
+\[P(n, k) = \frac{n!}{(n-k)!} = n(n-1)\cdots(n-k+1)\]
+
+**Combinations — order does not matter.** Choosing \(k\) of \(n\) with no regard to order:
+
+\[C(n, k) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!} = \frac{P(n,k)}{k!}\]
+
+A combination is a permutation with the \(k!\) internal orderings divided out. "Committee,"
+"group," "selection," "handshake" → combination. "Ranking," "arrangement," "line," "code" →
+permutation.
+
+**Identical items.** Arranging \(n\) items where some repeat (\(a\) alike, \(b\) alike, …):
+
+\[\frac{n!}{a!\,b!\,\cdots}\]
+
+Divide out the interchangeable duplicates. (This is why `LEVEL` has \(5!/(2!\,2!)\) arrangements,
+not \(5!\).)
+
+**Restrictions — two standard moves.**
+- *Must be together:* glue the group into one block, arrange the blocks, then multiply by the
+  internal arrangements of the block.
+- *Must be apart (no two adjacent):* seat the unrestricted items first, then drop the restricted
+  ones into the **gaps** between and around them.
+
+## Worked examples
+
+**Multiply.** A menu has 4 breads, 3 fillings, 2 spreads; one of each → \(4\times 3\times 2 = 24\)
+sandwiches.
+
+**Combination vs. permutation.** From 8 people, a *committee* of 3 is \(C(8,3) = 56\); a *ranked*
+top-3 is \(P(8,3) = 336\). The ratio is \(3! = 6\), the orderings a committee ignores.
+
+**Together.** Six people in a row with two specific people adjacent: treat the pair as one block →
+\(5!\) arrangements of the 5 units, times \(2!\) for the pair's order \(= 240\).
+
+## Common traps
+
+- **Adding when you should multiply.** Independent successive choices multiply; \(4+3+2\) is wrong
+  for the sandwich count.
+- **Permutation vs. combination.** Using \(P(n,k)\) for an unordered selection over-counts by
+  \(k!\); using \(C(n,k)\) for an ordered one under-counts.
+- **Forgetting to divide out identical items.** \(5!\) over-counts `LEVEL` because swapping the two
+  L's (or two E's) changes nothing.
+- **"At least one" head-on.** Counting cases one by one invites double-counting. Use the
+  complement: (total) − (none) is almost always cleaner.
+- **Restriction, then forget the internal order.** After gluing a block, remember to multiply by
+  the block's own arrangements.
+
+## Key takeaways
+
+- Decide multiply-vs-add and order-matters-vs-not *before* reaching for a formula.
+- \(C(n,k) = P(n,k)/k!\) — a combination is a permutation with the orderings removed.
+- Identical items: divide \(n!\) by the factorial of each repeat count.
+- "At least one" → subtract the complement from the total.
+- Restrictions: glue for "together," use gaps for "apart."

content/questions/quant/quant-counting-probability-combinatorics-0001.md

@@ -0,0 +1,45 @@
+---
+id: quant-counting-probability-combinatorics-0001
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: easy
+tags: [combinatorics, counting-principle]
+choices:
+  A: "8"
+  B: "9"
+  C: "12"
+  D: "24"
+  E: "48"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A café builds a sandwich by choosing exactly one of 4 breads, one of 3 cheeses, and one of 2
+spreads. How many distinct sandwiches can be made?
+
+## Explanation
+
+The three choices are made **together** (one bread *and* one cheese *and* one spread), so by the
+fundamental counting principle you **multiply**:
+
+\[4 \times 3 \times 2 = 24\]
+
+**Traps:**
+- **B (9)** *adds* the options \((4 + 3 + 2)\) — addition is for mutually exclusive cases ("or"),
+  not for independent successive choices ("and").
+- **C (12)** multiplies only bread and cheese \((4 \times 3)\), forgetting the spread.
+- **A (8)** multiplies only bread and spread \((4 \times 2)\), forgetting the cheese.
+- **E (48)** double-counts the spread \((4 \times 3 \times 2 \times 2)\).
+
+## Hints
+
+- Each independent choice made together multiplies the count — decide multiply vs. add first.
+- Include every stage: bread × cheese × spread.

content/questions/quant/quant-counting-probability-combinatorics-0002.md

@@ -0,0 +1,46 @@
+---
+id: quant-counting-probability-combinatorics-0002
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: easy
+tags: [combinatorics, permutations, factorial]
+choices:
+  A: "15"
+  B: "20"
+  C: "24"
+  D: "25"
+  E: "120"
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+In how many different ways can 5 distinct books be arranged in a row on a shelf?
+
+## Explanation
+
+Arranging \(n\) distinct items in a row is \(n!\). With 5 books:
+
+\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
+
+Think of it as filling positions left to right: 5 choices for the first slot, 4 for the next, then
+3, 2, 1 — multiplied together.
+
+**Traps:**
+- **A (15)** *adds* \(5 + 4 + 3 + 2 + 1\) instead of multiplying.
+- **B (20)** stops after two factors \((5 \times 4)\).
+- **C (24)** is \(4!\) — arranging only 4 of the books.
+- **D (25)** is \(5^2\), treating each slot as having all 5 books available (no removal after
+  placing one).
+
+## Hints
+
+- "Arrange all \(n\) distinct items in a row" is exactly \(n!\).
+- Each book used up reduces the choices for the next slot: \(5 \times 4 \times 3 \times 2 \times 1\).

content/questions/quant/quant-counting-probability-combinatorics-0003.md

@@ -0,0 +1,45 @@
+---
+id: quant-counting-probability-combinatorics-0003
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: medium
+tags: [combinatorics, combinations]
+choices:
+  A: "24"
+  B: "56"
+  C: "112"
+  D: "168"
+  E: "336"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A committee of 3 people is to be selected from a group of 8. How many different committees are
+possible?
+
+## Explanation
+
+A committee is an **unordered** selection — the same three people form the same committee no matter
+what order they're picked in — so this is a combination:
+
+\[C(8,3) = \frac{8!}{3!\,5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\]
+
+**Traps:**
+- **E (336)** is \(P(8,3) = 8 \times 7 \times 6\) — treating order as mattering, so each committee is
+  counted \(3! = 6\) times.
+- **C (112)** divides by 3 instead of \(3! = 6\) \((336/3)\).
+- **D (168)** divides by \(2!\) instead of \(3!\) \((336/2)\).
+- **A (24)** just computes \(8 \times 3\).
+
+## Hints
+
+- "Committee," "group," "selection" → order doesn't matter → combination.
+- A combination is the permutation \(P(n,k)\) divided by \(k!\) to remove the internal orderings.

content/questions/quant/quant-counting-probability-combinatorics-0004.md

@@ -0,0 +1,47 @@
+---
+id: quant-counting-probability-combinatorics-0004
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: medium
+tags: [combinatorics, permutations, identical-items]
+choices:
+  A: "10"
+  B: "20"
+  C: "30"
+  D: "60"
+  E: "120"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+How many distinct arrangements can be made of the letters in the word LEVEL?
+
+## Explanation
+
+LEVEL has 5 letters, but they are not all distinct: **L** appears twice, **E** appears twice, and
+**V** once. When items repeat, divide \(n!\) by the factorial of each repeat count:
+
+\[\frac{5!}{2!\,2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30\]
+
+The two L's are interchangeable (swapping them gives the same word), and likewise the two E's, so
+each distinct arrangement was counted \(2! \times 2! = 4\) times in the raw \(5!\).
+
+**Traps:**
+- **E (120)** is \(5!\) — treating all five letters as distinct.
+- **D (60)** divides by only one \(2!\) \((120/2)\), correcting for the L's but not the E's (or vice
+  versa).
+- **A (10)** over-divides, e.g. by an extra factor.
+- **B (20)** divides by \(3!\) \((120/6)\), as if three letters repeated.
+
+## Hints
+
+- Repeated, interchangeable letters mean the plain \(5!\) over-counts.
+- Divide \(5!\) by \(2!\) for the two L's **and** another \(2!\) for the two E's.

content/questions/quant/quant-counting-probability-combinatorics-0005.md

@@ -0,0 +1,49 @@
+---
+id: quant-counting-probability-combinatorics-0005
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: medium
+tags: [combinatorics, permutations, restrictions]
+choices:
+  A: "120"
+  B: "240"
+  C: "360"
+  D: "480"
+  E: "720"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Six friends are to be seated in a row of 6 chairs. If two particular friends insist on sitting
+next to each other, how many seating arrangements are possible?
+
+## Explanation
+
+Use the "glue" method for a must-be-together restriction. Treat the two friends as a single block,
+so you're now arranging **5 units** (the block plus the other 4 people):
+
+\[5! = 120 \text{ ways to arrange the units}\]
+
+The two friends inside the block can also swap seats, which is \(2! = 2\) arrangements:
+
+\[120 \times 2 = 240\]
+
+**Traps:**
+- **E (720)** is \(6!\) — ignoring the restriction entirely.
+- **A (120)** is \(5!\) — gluing the pair but forgetting they can be ordered two ways inside the
+  block.
+- **C (360)** halves \(6!\) \((720/2)\), a common but incorrect fix.
+- **D (480)** doubles the correct count, applying the internal \(\times 2\) one time too many.
+
+## Hints
+
+- "Must sit together" → glue the pair into one block and arrange the resulting units.
+- Don't forget the pair can be ordered two ways inside their block.