From 1048c0a43228f26f20089aa5cb4e62d9ff40cdf2 Mon Sep 17 00:00:00 2001 From: Claude Date: Sat, 25 Jul 2026 09:08:08 +0000 Subject: [PATCH] Add combinatorics bundle: lesson + 7 questions (quant/counting-probability) Covers the previously empty counting-probability/combinatorics subtopic with a concept lesson and 7 original questions in an easy->hard progression: fundamental counting principle, factorial arrangements, permutations (nPr), combinations (nCr), a restricted committee, letter arrangements with repeats, and circular seating. Each question has a worked explanation, two hints, and distractors keyed to specific common errors. Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_01FJ3gK9RbjNA17dZu2h2HCY --- ...uant-counting-probability-combinatorics.md | 111 +++++++++ ...counting-probability-combinatorics-0017.md | 47 ++++ ...counting-probability-combinatorics-0018.md | 44 ++++ ...counting-probability-combinatorics-0019.md | 50 ++++ ...counting-probability-combinatorics-0020.md | 46 ++++ ...counting-probability-combinatorics-0021.md | 50 ++++ ...counting-probability-combinatorics-0022.md | 47 ++++ ...counting-probability-combinatorics-0023.md | 48 ++++ docs/data/content.json | 234 +++++++++++++++++- 9 files changed, 673 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-counting-probability-combinatorics.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0017.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0018.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0019.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0020.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0021.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0022.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0023.md diff --git a/content/lessons/quant/quant-counting-probability-combinatorics.md b/content/lessons/quant/quant-counting-probability-combinatorics.md new file mode 100644 index 0000000..c323339 --- /dev/null +++ b/content/lessons/quant/quant-counting-probability-combinatorics.md @@ -0,0 +1,111 @@ +--- +id: quant-counting-probability-combinatorics +section: quant +topic: counting-probability +subtopic: combinatorics +title: "Combinatorics: Counting Without Listing" +tags: [combinatorics, counting, permutations, combinations, factorial] +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**. On GMAT Focus Quant the numbers are kept small on purpose — the exam is testing whether +you can spot the *structure* of a counting problem, not whether you can grind through a large +factorial. Almost every problem reduces to three questions you ask in order: + +1. Am I building the count from independent **stages** (pick a bread, then a cheese, then a drink)? +2. Does **order matter** among the things I select? +3. Are any of the items **identical**, so that some arrangements are really the same? + +Answer those three and the right formula falls out. + +## Core concepts + +**The fundamental counting principle.** If one choice can be made in \(m\) ways and an +independent second choice in \(n\) ways, the two together can be made in \(m \times n\) ways. +Stages **multiply**; they never add. Three stages with 5, 4, and 2 options give +\(5 \times 4 \times 2 = 40\) outcomes. + +**Factorial.** The number of ways to arrange \(n\) **distinct** items in a row is + +\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\] + +with the convention \(0! = 1\). So 5 distinct books fill a shelf in \(5! = 120\) orders. + +**Permutations — order matters.** Choosing \(k\) items from \(n\) and *arranging* them (distinct +roles, ranked positions, a seating order) is + +\[{}_nP_k = \frac{n!}{(n-k)!} = n \times (n-1) \times \cdots \times (n-k+1)\] + +Assigning president, vice-president, and treasurer to 8 people: \(8 \times 7 \times 6 = 336\). + +**Combinations — order does not matter.** Choosing \(k\) items from \(n\) as an unordered +*group* (a committee, a handful, a subset) is + +\[{}_nC_k = \binom{n}{k} = \frac{n!}{k!\,(n-k)!}\] + +The extra \(k!\) in the denominator divides out the \(k!\) orderings of the same group, which the +permutation count treated as different. A committee of 3 from 7 people: +\(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\). + +**A useful symmetry:** \(\binom{n}{k} = \binom{n}{n-k}\). Choosing which 5 of 7 to *include* is +the same as choosing which 2 to *leave out*. + +**Arrangements with identical items.** When some of the \(n\) items are indistinguishable, +divide \(n!\) by a factorial for each repeated group. The distinct arrangements of the letters in +a word with counts \(n_1, n_2, \dots\) are + +\[\frac{n!}{n_1!\,n_2!\cdots}\] + +**Circular arrangements.** Seating \(n\) people around a round table has only \((n-1)!\) distinct +arrangements, because a rotation of everyone is the same seating — fix one person and arrange the +rest. + +## Worked examples + +**Stages multiply.** A lunch special picks one of 5 mains, one of 4 sides, one of 2 drinks. The +three choices are independent, so multiply: \(5 \times 4 \times 2 = 40\) specials. Adding +(\(5+4+2 = 11\)) answers a different question — "how many single items are on the menu." + +**Order or not?** Pick 3 of 7 people. If you are ranking them 1st–2nd–3rd, order matters and the +count is \({}_7P_3 = 7 \times 6 \times 5 = 210\). If you are just forming a group of 3, order does +not matter and you divide by \(3! = 6\): \(\binom{7}{3} = 210 / 6 = 35\). Same selection, different +question — the word "committee," "group," or "set" signals a combination; "arrange," "order," +"rank," or distinct titles signal a permutation. + +**Split committees multiply.** Choose exactly 2 men from 5 **and** 2 women from 4: +\(\binom{5}{2} \times \binom{4}{2} = 10 \times 6 = 60\). The two selections are independent stages, +so the counts multiply — they are not added. + +**Identical letters.** The word BANANA has 6 letters: A appears 3 times, N twice, B once. Treating +all 6 as distinct overcounts, so divide by the repeats: +\(\frac{6!}{3!\,2!} = \frac{720}{12} = 60\). + +## Common traps + +- **Adding when you should multiply.** Independent stages multiply. Add only when you are counting + disjoint *cases* ("the number is red **or** blue"), never sequential choices. +- **Confusing permutations and combinations.** Distinct roles, rankings, or seatings ⇒ permutation + (order matters). An unordered group, committee, or handful ⇒ combination (divide by \(k!\)). +- **Dividing by the wrong factorial.** For a combination of \(k\) items you divide by \(k!\), not by + \(k\). \({}_7P_3 = 210\) becomes \(\binom{7}{3} = 210/3! = 35\), not \(210/3 = 70\). +- **Ignoring identical items.** Repeated letters or identical objects mean \(n!\) overcounts; divide + by a factorial for each repeated group, and **multiply** those factorials in the denominator — + never add them. +- **Forgetting a stage — or double-counting order in one.** In a split selection, count each group + as a combination unless that group is itself ordered. + +## Key takeaways + +- Independent choices **multiply** (the fundamental counting principle); disjoint cases add. +- Arranging \(n\) distinct items uses \(n!\); arranging \(k\) of \(n\) uses \({}_nP_k\). +- **Order matters ⇒ permutation; order does not ⇒ combination.** A combination is a permutation + divided by \(k!\). +- Identical items: divide \(n!\) by the factorial of each repeated group's count. +- Round table: \(n\) people seat in \((n-1)!\) ways — fix one seat to kill the rotations. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0017.md b/content/questions/quant/quant-counting-probability-combinatorics-0017.md new file mode 100644 index 0000000..58a74a0 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0017.md @@ -0,0 +1,47 @@ +--- +id: quant-counting-probability-combinatorics-0017 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: easy +tags: [combinatorics, fundamental-counting-principle] +choices: + A: "8" + B: "10" + C: "11" + D: "20" + E: "40" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A café's lunch special lets a customer choose exactly one of 5 main dishes, one of 4 side dishes, +and one of 2 drinks. How many different lunch specials can be assembled? + +## Explanation + +The three choices are made independently, one after another, so by the **fundamental counting +principle** you multiply the number of options at each stage: + +\[5 \times 4 \times 2 = 40\] + +Every one of the 5 mains can pair with every one of the 4 sides, and each of those 20 pairings can +pair with either of the 2 drinks — \(20 \times 2 = 40\). + +**The trap:** answer **C (11)** comes from *adding* \(5 + 4 + 2\), which counts how many individual +menu items exist, not how many complete specials can be built. Independent stages multiply; they +never add. The other wrong answers each drop a stage: **D (20)** is \(5 \times 4\) (forgetting the +drink), **B (10)** is \(5 \times 2\) (forgetting the side), and **A (8)** is \(4 \times 2\) +(forgetting the main). + +## Hints + +- Picking a main, then a side, then a drink is a sequence of independent choices. +- For independent stages, multiply the counts — don't add them. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0018.md b/content/questions/quant/quant-counting-probability-combinatorics-0018.md new file mode 100644 index 0000000..8f48791 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0018.md @@ -0,0 +1,44 @@ +--- +id: quant-counting-probability-combinatorics-0018 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: easy +tags: [combinatorics, factorial, arrangements] +choices: + A: "5" + B: "20" + C: "25" + D: "60" + E: "120" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +In how many different orders can 5 distinct books be arranged in a row on a shelf? + +## Explanation + +Arranging \(n\) **distinct** items in a row is a straight factorial count. There are 5 choices for +the leftmost slot, then 4 for the next (one book is already placed), then 3, then 2, then 1: + +\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\] + +**The trap:** answer **B (20)** stops after two factors (\(5 \times 4\)), as if only the first two +positions mattered — but every one of the 5 positions still needs to be filled. **C (25)** squares +the count (\(5^2\)), which would let a book repeat in multiple slots; here each book is used exactly +once. **D (60)** is \(5!/2\), an unwarranted division by 2 (there is no symmetry to remove — the +books are all different and the shelf has a fixed left-to-right order). **A (5)** just counts the +books, not their orderings. + +## Hints + +- Fill the positions one at a time: how many books can go first, then second, and so on? +- Arranging \(n\) distinct items in a row is \(n!\). diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0019.md b/content/questions/quant/quant-counting-probability-combinatorics-0019.md new file mode 100644 index 0000000..d97ecd6 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0019.md @@ -0,0 +1,50 @@ +--- +id: quant-counting-probability-combinatorics-0019 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: medium +tags: [combinatorics, permutations, order-matters] +choices: + A: "24" + B: "56" + C: "336" + D: "512" + E: "40320" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A club has 8 members. It must select a president, a vice-president, and a treasurer, and no member +may hold more than one of these positions. In how many different ways can the three positions be +filled? + +## Explanation + +The three positions are **distinct roles**, so order matters — being president is different from +being treasurer. Fill the roles one at a time: 8 people can be president, then 7 remain for +vice-president, then 6 for treasurer: + +\[8 \times 7 \times 6 = 336\] + +This is the permutation \({}_8P_3 = \dfrac{8!}{5!} = 336\). + +**The trap:** answer **B (56)** is \(\binom{8}{3}\), the number of ways to choose an *unordered* +group of 3 — but that ignores which of the three named jobs each person gets. Because the roles are +distinct, each group of 3 can be arranged in \(3! = 6\) ways, and \(56 \times 6 = 336\). **D (512)** +is \(8^3\), which would allow the same person to hold several positions (repetition), forbidden +here. **E (40320)** is \(8!\), arranging all 8 members. **A (24)** is \(8 \times 3\), a +multiplication that has no counting meaning. + +## Hints + +- The three positions are different jobs — does swapping two people's jobs create a new outcome? +- When order matters, use \({}_nP_k = n(n-1)\cdots\); with no repeats allowed, the pool shrinks by + one at each step. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0020.md b/content/questions/quant/quant-counting-probability-combinatorics-0020.md new file mode 100644 index 0000000..1e47456 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0020.md @@ -0,0 +1,46 @@ +--- +id: quant-counting-probability-combinatorics-0020 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: medium +tags: [combinatorics, combinations, order-does-not-matter] +choices: + A: "35" + B: "70" + C: "105" + D: "210" + E: "343" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +From a group of 7 people, how many different 3-person committees can be formed? (All committee +members serve in the same capacity — no titles or ranking.) + +## Explanation + +A committee is an **unordered** group: choosing Ann, Ben, then Cara gives the same committee as +choosing Cara, Ben, then Ann. So this is a combination: + +\[\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35\] + +**The trap:** answer **D (210)** is \(7 \times 6 \times 5 = {}_7P_3\) — it counts the *ordered* +selections, treating each rearrangement of the same three people as different. Since order does not +matter, you must divide by \(3! = 6\) to collapse those duplicates. **B (70)** divides by 3 instead +of \(3!\), and **C (105)** divides by 2 instead of \(3!\) — both are the "remembered to divide, but +by the wrong number" error. **E (343)** is \(7^3\), which allows repeats and order, neither of which +applies. + +## Hints + +- Does reordering the same three people create a different committee? +- A combination is the ordered count divided by \(k!\) — for a group of 3, divide by \(3! = 6\), not + by 3. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0021.md b/content/questions/quant/quant-counting-probability-combinatorics-0021.md new file mode 100644 index 0000000..c59da04 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0021.md @@ -0,0 +1,50 @@ +--- +id: quant-counting-probability-combinatorics-0021 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, combinations, restricted-selection] +choices: + A: "16" + B: "60" + C: "120" + D: "126" + E: "3024" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A 4-person committee is to be formed from a pool of 5 men and 4 women, and it must contain +**exactly 2 men and 2 women**. How many different committees are possible? + +## Explanation + +Handle the two requirements as independent stages and **multiply**. Choosing the men and choosing +the women are separate unordered selections: + +\[\binom{5}{2} \times \binom{4}{2} = \frac{5 \times 4}{2} \times \frac{4 \times 3}{2} = 10 \times 6 = 60\] + +Each of the 10 possible pairs of men can be teamed with each of the 6 possible pairs of women, so +the counts multiply. + +**The trap:** answer **A (16)** *adds* the two combinations (\(10 + 6\)) instead of multiplying — +but you need one pair from each group *together*, which is a product, not a sum. **D (126)** is +\(\binom{9}{4}\), the number of ways to pick any 4 of the 9 people while **ignoring** the 2-and-2 +restriction. **C (120)** counts the women as an *ordered* pair, using \({}_4P_2 = 12\) instead of +\(\binom{4}{2} = 6\) (\(10 \times 12 = 120\)) — but committee members have no order. **E (3024)** is +\({}_9P_4\), ordering all four choices and dropping the restriction entirely. + +## Hints + +- Split the committee into "pick the men" and "pick the women" — two separate choices. +- Each subgroup is unordered (a combination); then combine the two independent choices by + multiplying. + diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0022.md b/content/questions/quant/quant-counting-probability-combinatorics-0022.md new file mode 100644 index 0000000..8a36113 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0022.md @@ -0,0 +1,47 @@ +--- +id: quant-counting-probability-combinatorics-0022 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, arrangements, identical-items] +choices: + A: "60" + B: "90" + C: "120" + D: "360" + E: "720" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +How many distinct arrangements can be made of all the letters of the word **BANANA**? + +## Explanation + +BANANA has 6 letters, but they are not all distinct: **A** appears 3 times, **N** appears 2 times, +and **B** appears once. If the 6 letters were all different there would be \(6! = 720\) +arrangements — but swapping two identical A's (or the two N's) produces no new visible word, so +\(6!\) overcounts. Divide by a factorial for each repeated letter: + +\[\frac{6!}{3!\,\cdot\,2!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60\] + +**The trap:** answer **E (720)** is \(6!\), treating all six letters as distinguishable. **D (360)** +divides by only \(2!\) (correcting for the two N's but forgetting the three A's), and **C (120)** +divides by only \(3!\) (correcting for the A's but forgetting the N's). **B (90)** comes from +\(720 / (3! + 2!) = 720/8\) — *adding* the repeat-factorials in the denominator instead of +multiplying them. + +## Hints + +- All-distinct letters would give \(6!\); repeated letters make some of those arrangements + identical. +- Divide by the factorial of each repeated letter's count, and multiply those factorials together + in the denominator. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0023.md b/content/questions/quant/quant-counting-probability-combinatorics-0023.md new file mode 100644 index 0000000..a67d1c0 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0023.md @@ -0,0 +1,48 @@ +--- +id: quant-counting-probability-combinatorics-0023 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, circular-arrangements] +choices: + A: "6" + B: "36" + C: "60" + D: "120" + E: "720" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +Six friends are to be seated around a circular table with six chairs. Two seatings are considered +the same if one can be obtained from the other by rotating everyone around the table. How many +distinct seatings are possible? + +## Explanation + +Around a circle there is no fixed "first" seat, so the \(6!\) arrangements you would count for a +straight row overcount by a factor of 6 — every seating looks identical to the 5 other seatings you +get by rotating everyone one chair over. The clean way to handle this is to **fix one person's +seat** as a reference point and arrange the remaining 5 people relative to that person: + +\[(6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\] + +**The trap:** answer **E (720)** is \(6!\), the count for a *row* of labeled seats — it treats each +of the 6 rotations of the same circular seating as different. **C (60)** takes the row count and +further divides by 2 (\(6!/(6 \cdot 2)\)), correcting for reflections as well — appropriate for a +flippable object like a bracelet, but the problem only calls rotations equivalent, not mirror +images. **B (36)** is \(6^2\) and **A (6)** just counts the people, neither of which is an +arrangement count. + +## Hints + +- On a circle there is no fixed starting seat, so rotating everyone gives the same seating. +- Anchor one person's chair and arrange the other \(n-1\) around them: \((n-1)!\). diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..acbc5a0 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 23, + "lessons": 6 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 39, + "lessons": 9 }, "questions": [ { @@ -359,6 +359,214 @@ "Multiply the factors together and compare to 1." ] }, + { + "id": "quant-counting-probability-combinatorics-0017", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "combinatorics", + "fundamental-counting-principle" + ], + "choices": { + "A": "8", + "B": "10", + "C": "11", + "D": "20", + "E": "40" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A café's lunch special lets a customer choose exactly one of 5 main dishes, one of 4 side dishes,\nand one of 2 drinks. How many different lunch specials can be assembled?", + "explanation": "The three choices are made independently, one after another, so by the **fundamental counting\nprinciple** you multiply the number of options at each stage:\n\n\\[5 \\times 4 \\times 2 = 40\\]\n\nEvery one of the 5 mains can pair with every one of the 4 sides, and each of those 20 pairings can\npair with either of the 2 drinks — \\(20 \\times 2 = 40\\).\n\n**The trap:** answer **C (11)** comes from *adding* \\(5 + 4 + 2\\), which counts how many individual\nmenu items exist, not how many complete specials can be built. Independent stages multiply; they\nnever add. The other wrong answers each drop a stage: **D (20)** is \\(5 \\times 4\\) (forgetting the\ndrink), **B (10)** is \\(5 \\times 2\\) (forgetting the side), and **A (8)** is \\(4 \\times 2\\)\n(forgetting the main).", + "hints": [ + "Picking a main, then a side, then a drink is a sequence of independent choices.", + "For independent stages, multiply the counts — don't add them." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0018", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "combinatorics", + "factorial", + "arrangements" + ], + "choices": { + "A": "5", + "B": "20", + "C": "25", + "D": "60", + "E": "120" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "In how many different orders can 5 distinct books be arranged in a row on a shelf?", + "explanation": "Arranging \\(n\\) **distinct** items in a row is a straight factorial count. There are 5 choices for\nthe leftmost slot, then 4 for the next (one book is already placed), then 3, then 2, then 1:\n\n\\[5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120\\]\n\n**The trap:** answer **B (20)** stops after two factors (\\(5 \\times 4\\)), as if only the first two\npositions mattered — but every one of the 5 positions still needs to be filled. **C (25)** squares\nthe count (\\(5^2\\)), which would let a book repeat in multiple slots; here each book is used exactly\nonce. **D (60)** is \\(5!/2\\), an unwarranted division by 2 (there is no symmetry to remove — the\nbooks are all different and the shelf has a fixed left-to-right order). **A (5)** just counts the\nbooks, not their orderings.", + "hints": [ + "Fill the positions one at a time: how many books can go first, then second, and so on?", + "Arranging \\(n\\) distinct items in a row is \\(n!\\)." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0019", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "combinatorics", + "permutations", + "order-matters" + ], + "choices": { + "A": "24", + "B": "56", + "C": "336", + "D": "512", + "E": "40320" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A club has 8 members. It must select a president, a vice-president, and a treasurer, and no member\nmay hold more than one of these positions. In how many different ways can the three positions be\nfilled?", + "explanation": "The three positions are **distinct roles**, so order matters — being president is different from\nbeing treasurer. Fill the roles one at a time: 8 people can be president, then 7 remain for\nvice-president, then 6 for treasurer:\n\n\\[8 \\times 7 \\times 6 = 336\\]\n\nThis is the permutation \\({}_8P_3 = \\dfrac{8!}{5!} = 336\\).\n\n**The trap:** answer **B (56)** is \\(\\binom{8}{3}\\), the number of ways to choose an *unordered*\ngroup of 3 — but that ignores which of the three named jobs each person gets. Because the roles are\ndistinct, each group of 3 can be arranged in \\(3! = 6\\) ways, and \\(56 \\times 6 = 336\\). **D (512)**\nis \\(8^3\\), which would allow the same person to hold several positions (repetition), forbidden\nhere. **E (40320)** is \\(8!\\), arranging all 8 members. **A (24)** is \\(8 \\times 3\\), a\nmultiplication that has no counting meaning.", + "hints": [ + "The three positions are different jobs — does swapping two people's jobs create a new outcome?", + "When order matters, use \\({}_nP_k = n(n-1)\\cdots\\); with no repeats allowed, the pool shrinks by" + ] + }, + { + "id": "quant-counting-probability-combinatorics-0020", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "combinatorics", + "combinations", + "order-does-not-matter" + ], + "choices": { + "A": "35", + "B": "70", + "C": "105", + "D": "210", + "E": "343" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "From a group of 7 people, how many different 3-person committees can be formed? (All committee\nmembers serve in the same capacity — no titles or ranking.)", + "explanation": "A committee is an **unordered** group: choosing Ann, Ben, then Cara gives the same committee as\nchoosing Cara, Ben, then Ann. So this is a combination:\n\n\\[\\binom{7}{3} = \\frac{7 \\times 6 \\times 5}{3 \\times 2 \\times 1} = \\frac{210}{6} = 35\\]\n\n**The trap:** answer **D (210)** is \\(7 \\times 6 \\times 5 = {}_7P_3\\) — it counts the *ordered*\nselections, treating each rearrangement of the same three people as different. Since order does not\nmatter, you must divide by \\(3! = 6\\) to collapse those duplicates. **B (70)** divides by 3 instead\nof \\(3!\\), and **C (105)** divides by 2 instead of \\(3!\\) — both are the \"remembered to divide, but\nby the wrong number\" error. **E (343)** is \\(7^3\\), which allows repeats and order, neither of which\napplies.", + "hints": [ + "Does reordering the same three people create a different committee?", + "A combination is the ordered count divided by \\(k!\\) — for a group of 3, divide by \\(3! = 6\\), not" + ] + }, + { + "id": "quant-counting-probability-combinatorics-0021", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "combinations", + "restricted-selection" + ], + "choices": { + "A": "16", + "B": "60", + "C": "120", + "D": "126", + "E": "3024" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A 4-person committee is to be formed from a pool of 5 men and 4 women, and it must contain\n**exactly 2 men and 2 women**. How many different committees are possible?", + "explanation": "Handle the two requirements as independent stages and **multiply**. Choosing the men and choosing\nthe women are separate unordered selections:\n\n\\[\\binom{5}{2} \\times \\binom{4}{2} = \\frac{5 \\times 4}{2} \\times \\frac{4 \\times 3}{2} = 10 \\times 6 = 60\\]\n\nEach of the 10 possible pairs of men can be teamed with each of the 6 possible pairs of women, so\nthe counts multiply.\n\n**The trap:** answer **A (16)** *adds* the two combinations (\\(10 + 6\\)) instead of multiplying —\nbut you need one pair from each group *together*, which is a product, not a sum. **D (126)** is\n\\(\\binom{9}{4}\\), the number of ways to pick any 4 of the 9 people while **ignoring** the 2-and-2\nrestriction. **C (120)** counts the women as an *ordered* pair, using \\({}_4P_2 = 12\\) instead of\n\\(\\binom{4}{2} = 6\\) (\\(10 \\times 12 = 120\\)) — but committee members have no order. **E (3024)** is\n\\({}_9P_4\\), ordering all four choices and dropping the restriction entirely.", + "hints": [ + "Split the committee into \"pick the men\" and \"pick the women\" — two separate choices.", + "Each subgroup is unordered (a combination); then combine the two independent choices by" + ] + }, + { + "id": "quant-counting-probability-combinatorics-0022", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "arrangements", + "identical-items" + ], + "choices": { + "A": "60", + "B": "90", + "C": "120", + "D": "360", + "E": "720" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "How many distinct arrangements can be made of all the letters of the word **BANANA**?", + "explanation": "BANANA has 6 letters, but they are not all distinct: **A** appears 3 times, **N** appears 2 times,\nand **B** appears once. If the 6 letters were all different there would be \\(6! = 720\\)\narrangements — but swapping two identical A's (or the two N's) produces no new visible word, so\n\\(6!\\) overcounts. Divide by a factorial for each repeated letter:\n\n\\[\\frac{6!}{3!\\,\\cdot\\,2!} = \\frac{720}{6 \\times 2} = \\frac{720}{12} = 60\\]\n\n**The trap:** answer **E (720)** is \\(6!\\), treating all six letters as distinguishable. **D (360)**\ndivides by only \\(2!\\) (correcting for the two N's but forgetting the three A's), and **C (120)**\ndivides by only \\(3!\\) (correcting for the A's but forgetting the N's). **B (90)** comes from\n\\(720 / (3! + 2!) = 720/8\\) — *adding* the repeat-factorials in the denominator instead of\nmultiplying them.", + "hints": [ + "All-distinct letters would give \\(6!\\); repeated letters make some of those arrangements", + "Divide by the factorial of each repeated letter's count, and multiply those factorials together" + ] + }, + { + "id": "quant-counting-probability-combinatorics-0023", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "circular-arrangements" + ], + "choices": { + "A": "6", + "B": "36", + "C": "60", + "D": "120", + "E": "720" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "Six friends are to be seated around a circular table with six chairs. Two seatings are considered\nthe same if one can be obtained from the other by rotating everyone around the table. How many\ndistinct seatings are possible?", + "explanation": "Around a circle there is no fixed \"first\" seat, so the \\(6!\\) arrangements you would count for a\nstraight row overcount by a factor of 6 — every seating looks identical to the 5 other seatings you\nget by rotating everyone one chair over. The clean way to handle this is to **fix one person's\nseat** as a reference point and arrange the remaining 5 people relative to that person:\n\n\\[(6-1)! = 5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120\\]\n\n**The trap:** answer **E (720)** is \\(6!\\), the count for a *row* of labeled seats — it treats each\nof the 6 rotations of the same circular seating as different. **C (60)** takes the row count and\nfurther divides by 2 (\\(6!/(6 \\cdot 2)\\)), correcting for reflections as well — appropriate for a\nflippable object like a bracelet, but the problem only calls rotations equivalent, not mirror\nimages. **B (36)** is \\(6^2\\) and **A (6)** just counts the people, neither of which is an\narrangement count.", + "hints": [ + "On a circle there is no fixed starting seat, so rotating everyone gives the same seating.", + "Anchor one person's chair and arrange the other \\(n-1\\) around them: \\((n-1)!\\)." + ] + }, { "id": "quant-counting-probability-probability-0007", "section": "quant", @@ -984,6 +1192,24 @@ "status": "in-review", "body": "## Overview\n\nPercents and ratios are the most frequently tested arithmetic ideas on GMAT Focus Quant, and\nthey hide inside word problems everywhere. Master three moves: converting between forms, handling\npercent *change*, and scaling ratios.\n\n## Core concepts\n\n**Percent as a factor.** A percent is just a number over 100. The fastest way to apply a percent\nchange is to turn it into a **multiplier**:\n\n- Increase by \\(r\\%\\): multiply by \\(1 + \\tfrac{r}{100}\\). (+25% → \\(\\times 1.25\\))\n- Decrease by \\(r\\%\\): multiply by \\(1 - \\tfrac{r}{100}\\). (−20% → \\(\\times 0.80\\))\n\n**Successive changes multiply.** Two changes in a row are multiplied, never added:\n\n\\[(+25\\%)\\text{ then }(-20\\%): \\quad 1.25 \\times 0.80 = 1.00 \\quad (\\text{no net change})\\]\n\n**Percent change formula.**\n\n\\[\\text{percent change} = \\frac{\\text{new} - \\text{old}}{\\text{old}} \\times 100\\%\\]\n\nAlways divide by the **original** value, not the new one.\n\n**Ratios scale together.** A ratio \\(a : b\\) means the quantities are \\(ak\\) and \\(bk\\) for some\nmultiplier \\(k\\). If boys : girls \\(= 3 : 5\\) and there are 24 boys, then \\(k = 8\\), so there are\n\\(5 \\times 8 = 40\\) girls.\n\n## Worked examples\n\n**Percent change.** A stock rises from $80 to $100. Percent increase \\(= \\tfrac{100 - 80}{80} = \\tfrac{20}{80} = 25\\%\\).\nNote it later falls from $100 back to $80: that's \\(\\tfrac{-20}{100} = -20\\%\\) — a *smaller* percent, because the base is now larger.\n\n**Ratio scaling.** A recipe uses flour : sugar \\(= 7 : 2\\). To use 21 cups of flour, \\(k = 3\\), so\nyou need \\(2 \\times 3 = 6\\) cups of sugar.\n\n## Common traps\n\n- **Adding successive percents.** +25% then −20% is *not* +5%; it's \\(1.25 \\times 0.80 = 1.00\\).\n- **Wrong base.** Percent change always divides by the original amount. A rise then an equal-percent fall does not return to the start.\n- **Ratio ≠ actual count.** \\(3 : 5\\) does not mean 3 and 5 — it means \\(3k\\) and \\(5k\\). Find \\(k\\) first.\n\n## Key takeaways\n\n- Convert percent changes to multipliers and multiply them for successive changes.\n- Percent change = (new − old) / old.\n- A ratio \\(a : b\\) represents \\(ak\\) and \\(bk\\); solve for the multiplier \\(k\\), then scale." }, + { + "id": "quant-counting-probability-combinatorics", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "title": "Combinatorics: Counting Without Listing", + "tags": [ + "combinatorics", + "counting", + "permutations", + "combinations", + "factorial" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nCombinatorics is the art of counting arrangements and selections **without writing them all\nout**. On GMAT Focus Quant the numbers are kept small on purpose — the exam is testing whether\nyou can spot the *structure* of a counting problem, not whether you can grind through a large\nfactorial. Almost every problem reduces to three questions you ask in order:\n\n1. Am I building the count from independent **stages** (pick a bread, then a cheese, then a drink)?\n2. Does **order matter** among the things I select?\n3. Are any of the items **identical**, so that some arrangements are really the same?\n\nAnswer those three and the right formula falls out.\n\n## Core concepts\n\n**The fundamental counting principle.** If one choice can be made in \\(m\\) ways and an\nindependent second choice in \\(n\\) ways, the two together can be made in \\(m \\times n\\) ways.\nStages **multiply**; they never add. Three stages with 5, 4, and 2 options give\n\\(5 \\times 4 \\times 2 = 40\\) outcomes.\n\n**Factorial.** The number of ways to arrange \\(n\\) **distinct** items in a row is\n\n\\[n! = n \\times (n-1) \\times (n-2) \\times \\cdots \\times 2 \\times 1\\]\n\nwith the convention \\(0! = 1\\). So 5 distinct books fill a shelf in \\(5! = 120\\) orders.\n\n**Permutations — order matters.** Choosing \\(k\\) items from \\(n\\) and *arranging* them (distinct\nroles, ranked positions, a seating order) is\n\n\\[{}_nP_k = \\frac{n!}{(n-k)!} = n \\times (n-1) \\times \\cdots \\times (n-k+1)\\]\n\nAssigning president, vice-president, and treasurer to 8 people: \\(8 \\times 7 \\times 6 = 336\\).\n\n**Combinations — order does not matter.** Choosing \\(k\\) items from \\(n\\) as an unordered\n*group* (a committee, a handful, a subset) is\n\n\\[{}_nC_k = \\binom{n}{k} = \\frac{n!}{k!\\,(n-k)!}\\]\n\nThe extra \\(k!\\) in the denominator divides out the \\(k!\\) orderings of the same group, which the\npermutation count treated as different. A committee of 3 from 7 people:\n\\(\\binom{7}{3} = \\frac{7 \\times 6 \\times 5}{3 \\times 2 \\times 1} = 35\\).\n\n**A useful symmetry:** \\(\\binom{n}{k} = \\binom{n}{n-k}\\). Choosing which 5 of 7 to *include* is\nthe same as choosing which 2 to *leave out*.\n\n**Arrangements with identical items.** When some of the \\(n\\) items are indistinguishable,\ndivide \\(n!\\) by a factorial for each repeated group. The distinct arrangements of the letters in\na word with counts \\(n_1, n_2, \\dots\\) are\n\n\\[\\frac{n!}{n_1!\\,n_2!\\cdots}\\]\n\n**Circular arrangements.** Seating \\(n\\) people around a round table has only \\((n-1)!\\) distinct\narrangements, because a rotation of everyone is the same seating — fix one person and arrange the\nrest.\n\n## Worked examples\n\n**Stages multiply.** A lunch special picks one of 5 mains, one of 4 sides, one of 2 drinks. The\nthree choices are independent, so multiply: \\(5 \\times 4 \\times 2 = 40\\) specials. Adding\n(\\(5+4+2 = 11\\)) answers a different question — \"how many single items are on the menu.\"\n\n**Order or not?** Pick 3 of 7 people. If you are ranking them 1st–2nd–3rd, order matters and the\ncount is \\({}_7P_3 = 7 \\times 6 \\times 5 = 210\\). If you are just forming a group of 3, order does\nnot matter and you divide by \\(3! = 6\\): \\(\\binom{7}{3} = 210 / 6 = 35\\). Same selection, different\nquestion — the word \"committee,\" \"group,\" or \"set\" signals a combination; \"arrange,\" \"order,\"\n\"rank,\" or distinct titles signal a permutation.\n\n**Split committees multiply.** Choose exactly 2 men from 5 **and** 2 women from 4:\n\\(\\binom{5}{2} \\times \\binom{4}{2} = 10 \\times 6 = 60\\). The two selections are independent stages,\nso the counts multiply — they are not added.\n\n**Identical letters.** The word BANANA has 6 letters: A appears 3 times, N twice, B once. Treating\nall 6 as distinct overcounts, so divide by the repeats:\n\\(\\frac{6!}{3!\\,2!} = \\frac{720}{12} = 60\\).\n\n## Common traps\n\n- **Adding when you should multiply.** Independent stages multiply. Add only when you are counting\n disjoint *cases* (\"the number is red **or** blue\"), never sequential choices.\n- **Confusing permutations and combinations.** Distinct roles, rankings, or seatings ⇒ permutation\n (order matters). An unordered group, committee, or handful ⇒ combination (divide by \\(k!\\)).\n- **Dividing by the wrong factorial.** For a combination of \\(k\\) items you divide by \\(k!\\), not by\n \\(k\\). \\({}_7P_3 = 210\\) becomes \\(\\binom{7}{3} = 210/3! = 35\\), not \\(210/3 = 70\\).\n- **Ignoring identical items.** Repeated letters or identical objects mean \\(n!\\) overcounts; divide\n by a factorial for each repeated group, and **multiply** those factorials in the denominator —\n never add them.\n- **Forgetting a stage — or double-counting order in one.** In a split selection, count each group\n as a combination unless that group is itself ordered.\n\n## Key takeaways\n\n- Independent choices **multiply** (the fundamental counting principle); disjoint cases add.\n- Arranging \\(n\\) distinct items uses \\(n!\\); arranging \\(k\\) of \\(n\\) uses \\({}_nP_k\\).\n- **Order matters ⇒ permutation; order does not ⇒ combination.** A combination is a permutation\n divided by \\(k!\\).\n- Identical items: divide \\(n!\\) by the factorial of each repeated group's count.\n- Round table: \\(n\\) people seat in \\((n-1)!\\) ways — fix one seat to kill the rotations." + }, { "id": "quant-statistics-descriptive", "section": "quant",