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..681e3b0 --- /dev/null +++ b/content/lessons/quant/quant-counting-probability-combinatorics.md @@ -0,0 +1,89 @@ +--- +id: quant-counting-probability-combinatorics +section: quant +topic: counting-probability +subtopic: combinatorics +title: "Combinatorics: Counting Arrangements & Selections" +tags: [combinatorics, counting-principle, permutations, combinations, factorial] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Combinatorics is the art of counting *how many* ways something can happen without listing them +all. On GMAT Focus Quant the numbers stay small, so the challenge is never the arithmetic — it is +deciding **which counting move applies**. Three questions sort out almost everything: + +1. Am I making a sequence of independent choices? → **multiply** (fundamental counting principle). +2. Am I arranging items where **order matters**? → **permutation**. +3. Am I selecting a group where **order does not matter**? → **combination**. + +Get the classification right and the formula is easy. + +## Core concepts + +**Fundamental counting principle.** If one stage can happen in \(m\) ways and a second, independent +stage in \(n\) ways, the two together happen in \(m \times n\) ways. This extends to any number of +stages — you **multiply**, never add. (You add only when the choices are mutually exclusive +alternatives — "either this *or* that" — not sequential steps.) + +**Factorial.** \(n! = n \times (n-1) \times \cdots \times 2 \times 1\), the number of ways to arrange +\(n\) distinct items in a row. By convention \(0! = 1\). + +**Permutations — order matters.** The number of ways to arrange \(k\) of \(n\) distinct items in +order: + +\[P(n,k) = \frac{n!}{(n-k)!} = n \times (n-1) \times \cdots \times (n-k+1)\] + +Example: 3 of 5 books in order = \(5 \times 4 \times 3 = 60\). + +**Combinations — order does not matter.** The number of ways to *select* \(k\) of \(n\) distinct +items when arrangement is irrelevant: + +\[C(n,k) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!} = \frac{P(n,k)}{k!}\] + +A combination is just a permutation with the internal orderings divided out. Example: a committee of +3 from 8 people = \(\dfrac{8 \times 7 \times 6}{3!} = \dfrac{336}{6} = 56\). + +Two handy facts: \(C(n,k) = C(n,\,n-k)\) (choosing who is *in* is the same as choosing who is *out*), +and \(C(n,0) = C(n,n) = 1\). + +## Worked examples + +**Multiply the stages.** A meal is one appetizer (4 options), one entrée (5), and one dessert (3). +Meals \(= 4 \times 5 \times 3 = 60\). Independent, sequential choices → multiply. + +**Arrange with a restriction.** In how many ways can 5 people sit in a row if two specific people +must sit together? Glue the pair into a single block: now 4 units arrange in \(4! = 24\) ways, and the +pair can switch places inside the block in \(2! = 2\) ways, giving \(24 \times 2 = 48\). + +**Split into groups.** Choose a committee of 4 from 5 men and 4 women with exactly 2 of each. Choose +the men *and* the women, so multiply: \(C(5,2) \times C(4,2) = 10 \times 6 = 60\). + +**"At least" → count the complement.** From 9 people, how many teams of 4 include at least one of the +3 juniors? Counting "at least one" directly means adding several cases; it is faster to take *all* +teams minus the teams with *no* junior: \(C(9,4) - C(6,4) = 126 - 15 = 111\). + +## Common traps + +- **Adding when you should multiply.** Sequential independent stages multiply. Adding gives a number + far too small. +- **Permutation vs. combination.** If the answer to "does rearranging the same items give a *different* + outcome?" is no, use a combination. A committee of {A, B, C} is the same committee as {C, B, A} — do + not count the orderings. Using \(P(n,k)\) for a selection over-counts by a factor of \(k!\). +- **Forgetting the internal order of a block.** When you glue items together for an + "adjacent" restriction, remember to multiply by the arrangements *inside* the block. +- **Attacking "at least one" head-on.** Complementary counting (total − none) is almost always cleaner + and avoids double-counting the overlap cases. + +## Key takeaways + +- Independent, sequential choices **multiply**; mutually exclusive alternatives add. +- **Order matters → permutation** \(\left(\tfrac{n!}{(n-k)!}\right)\); **order doesn't → combination** + \(\left(\tfrac{n!}{k!(n-k)!}\right)\). A combination is a permutation divided by \(k!\). +- Break "and" problems into stages and multiply the counts for each stage. +- For "at least one," count the total and subtract the "none" case. 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..3d7a477 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0017.md @@ -0,0 +1,50 @@ +--- +id: quant-counting-probability-combinatorics-0017 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: easy +tags: [combinatorics, counting-principle, fundamental-counting] +choices: + A: "8" + B: "9" + C: "12" + D: "14" + E: "24" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A café builds a sandwich by choosing exactly one bread (3 kinds), one filling (4 kinds), and one +sauce (2 kinds). How many different sandwiches can be made? + +## Explanation + +The three choices are made in sequence and are independent of one another, so by the **fundamental +counting principle** you **multiply** the number of options at each stage: + +\[3 \times 4 \times 2 = 24\] + +**The traps:** + +- **B (9)** adds the options \((3 + 4 + 2)\) instead of multiplying. You add only when the choices are + mutually exclusive alternatives; here you make one choice from *each* category, so it's "and," which + means multiply. +- **C (12)** multiplies bread × filling \((3 \times 4)\) but forgets the sauce entirely. +- **A (8)** multiplies filling × sauce \((4 \times 2)\) and drops the bread. +- **D (14)** multiplies two categories and then *adds* the third \((3 \times 4 + 2)\), mixing the + operations. + +Only **E (24)** multiplies all three. + +## Hints + +- Each sandwich needs one bread AND one filling AND one sauce — "and" between independent choices means multiply. +- Multiply the count at every stage; don't leave a category out. 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..73177c9 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0018.md @@ -0,0 +1,48 @@ +--- +id: quant-counting-probability-combinatorics-0018 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: easy +tags: [combinatorics, permutations, factorial, arrangements] +choices: + A: "15" + 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 ways can 5 distinct books be arranged in a row on a shelf? + +## Explanation + +Arranging all 5 distinct books in order is a **permutation of 5 items**: the first position can be +any of 5 books, the next any of the remaining 4, then 3, then 2, then 1: + +\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\] + +**The traps:** + +- **A (15)** *adds* the descending counts \((5 + 4 + 3 + 2 + 1)\) instead of multiplying them. +- **B (20)** stops after two positions \((5 \times 4)\) — that counts arrangements of only 2 of the + books, not all 5. +- **D (60)** stops after three positions \((5 \times 4 \times 3)\), arranging only 3 books. +- **C (25)** computes \(5^2\), as if each slot could independently be any of 5 books; but once a book + is placed it can't be reused, so the pool shrinks each step. + +Only **E (120)** multiplies all the way down to 1. + +## Hints + +- Once a book is placed in a spot, it can't go in another — so the number of choices drops by one at each position. +- Arranging all \(n\) 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..027aba0 --- /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, combinations, committee, order-does-not-matter] +choices: + A: "21" + B: "24" + C: "56" + D: "168" + E: "336" +answer: C +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 people. How many different committees +are possible? + +## Explanation + +A committee is a **selection where order does not matter** — the committee {A, B, C} is the same as +{C, B, A}. That's a **combination**: + +\[C(8,3) = \frac{8 \times 7 \times 6}{3!} = \frac{336}{6} = 56\] + +**The traps:** + +- **E (336)** computes \(8 \times 7 \times 6\), a *permutation* — it counts every ordering of the same + three people as different. Since order doesn't matter for a committee, you must divide by the + \(3! = 6\) ways to arrange each trio. +- **D (168)** starts from the 336 orderings but divides by 2 instead of by \(3! = 6\). Three chosen + people can be arranged in 6 ways, not 2. +- **A (21)** adds \(8 + 7 + 6\); **B (24)** multiplies \(8 \times 3\). Neither reflects choosing 3 + distinct members from the group. + +Only **C (56)** correctly divides out the ordering. + +## Hints + +- Does rearranging the same three people give a *different* committee? If no, it's a combination, not a permutation. +- Take the ordered count \(8 \times 7 \times 6\) and divide by \(3!\) to remove the duplicate arrangements. 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..98695c6 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0020.md @@ -0,0 +1,52 @@ +--- +id: quant-counting-probability-combinatorics-0020 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: medium +tags: [combinatorics, combinations, groups, and-rule] +choices: + A: "10" + B: "16" + C: "60" + D: "126" + E: "240" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A committee of 4 people is formed from 5 men and 4 women. How many committees consist of exactly +2 men and 2 women? + +## Explanation + +Build the committee in two independent stages — choose the men **and** choose the women — then +multiply the counts (order doesn't matter within either group, so use combinations): + +\[C(5,2) \times C(4,2) = 10 \times 6 = 60\] + +Here \(C(5,2) = \dfrac{5 \times 4}{2} = 10\) and \(C(4,2) = \dfrac{4 \times 3}{2} = 6\). + +**The traps:** + +- **B (16)** *adds* the two counts \((10 + 6)\) instead of multiplying. You need 2 men *and* 2 women, + so the stages multiply. +- **A (10)** stops after choosing the men \((C(5,2))\) and forgets to choose the women. +- **D (126)** ignores the "exactly 2 and 2" requirement and just picks any 4 of the 9 people + \((C(9,4) = 126)\). +- **E (240)** treats the picks as ordered, using permutations \(P(5,2) \times P(4,2) = 20 \times 12\); + but the two men on a committee aren't ranked, so order shouldn't be counted. + +Only **C (60)** multiplies the two combinations. + +## Hints + +- "Exactly 2 men and 2 women" is a choice from each group — choose men AND women, then multiply. +- Within each group the members aren't ranked, so use combinations, and \(C(5,2)=10\), \(C(4,2)=6\). 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..a058b3b --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0021.md @@ -0,0 +1,54 @@ +--- +id: quant-counting-probability-combinatorics-0021 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, permutations, arrangements, restriction, block-method] +choices: + A: "24" + B: "48" + C: "72" + D: "120" + E: "240" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +Five friends are seated in a row of 5 chairs. In how many of the possible seating arrangements +are two particular friends, Ana and Ben, seated next to each other? + +## Explanation + +Use the **block method**: since Ana and Ben must be adjacent, glue them into a single "super-person." +That leaves **4 units** to arrange in a row: + +\[4! = 24 \text{ ways.}\] + +But inside the block, Ana and Ben can sit in either order (Ana–Ben or Ben–Ana), which doubles the +count: + +\[4! \times 2 = 24 \times 2 = 48\] + +**The traps:** + +- **A (24)** treats the pair as one block \((4!)\) but forgets that Ana and Ben can swap seats within + the block — you must multiply by \(2\). +- **D (120)** is \(5!\), the total number of arrangements with **no** restriction. +- **C (72)** is the number of arrangements in which Ana and Ben are **not** together + \((120 - 48)\) — the complement of what's asked. +- **E (240)** doubles the full \(5!\) \((2 \times 120)\), double-counting rather than restricting. + +Only **B (48)** applies the block method correctly. + +## Hints + +- Force the two friends together by treating them as a single block, then arrange the resulting units. +- Don't forget the two people inside the block can still switch places with each other. 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..f0c1a22 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0022.md @@ -0,0 +1,56 @@ +--- +id: quant-counting-probability-combinatorics-0022 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, combinations, at-least-one, complementary-counting] +choices: + A: "15" + B: "60" + C: "111" + D: "126" + E: "168" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A team of 4 is to be chosen from a group of 6 seniors and 3 juniors (9 people in all). How many +possible teams include **at least one** junior? + +## Explanation + +"At least one" invites **complementary counting**: count *all* teams, then subtract the teams that +have **no** junior (i.e., all seniors). + +- Total teams of 4 from 9: \(C(9,4) = \dfrac{9 \times 8 \times 7 \times 6}{4!} = \dfrac{3024}{24} = 126\). +- Teams with no junior (all 4 from the 6 seniors): \(C(6,4) = C(6,2) = \dfrac{6 \times 5}{2} = 15\). + +\[\text{At least one junior} = 126 - 15 = 111\] + +**The traps:** + +- **D (126)** is the total number of teams — it forgets to remove the all-senior teams. +- **A (15)** is the number of teams with **no** junior — the complement that should be *subtracted*, + not reported. +- **E (168)** comes from "pick 1 junior to guarantee, then any 3 of the remaining 8": + \(C(3,1) \times C(8,3) = 3 \times 56 = 168\). This **over-counts** teams that have 2 or 3 juniors, + because each such team gets counted once for each junior that could have been the "guaranteed" one. +- **B (60)** counts only the teams with **exactly one** junior \((C(3,1) \times C(6,3) = 3 \times 20)\), + missing teams with 2 or 3 juniors. + +Only **C (111)** counts every team with at least one junior exactly once. (Check: +exactly-1 \(=60\), exactly-2 \(= C(3,2)C(6,2) = 3 \times 15 = 45\), exactly-3 \(= C(3,3)C(6,1) = 6\); +\(60 + 45 + 6 = 111\).) + +## Hints + +- "At least one" is usually fastest as total minus none — count all teams, then subtract the teams with zero juniors. +- The "zero juniors" case means all 4 members come from the 6 seniors. diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..108a28e 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 22, + "lessons": 6 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 38, + "lessons": 9 }, "questions": [ { @@ -359,6 +359,192 @@ "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", + "counting-principle", + "fundamental-counting" + ], + "choices": { + "A": "8", + "B": "9", + "C": "12", + "D": "14", + "E": "24" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A café builds a sandwich by choosing exactly one bread (3 kinds), one filling (4 kinds), and one\nsauce (2 kinds). How many different sandwiches can be made?", + "explanation": "The three choices are made in sequence and are independent of one another, so by the **fundamental\ncounting principle** you **multiply** the number of options at each stage:\n\n\\[3 \\times 4 \\times 2 = 24\\]\n\n**The traps:**\n\n- **B (9)** adds the options \\((3 + 4 + 2)\\) instead of multiplying. You add only when the choices are\n mutually exclusive alternatives; here you make one choice from *each* category, so it's \"and,\" which\n means multiply.\n- **C (12)** multiplies bread × filling \\((3 \\times 4)\\) but forgets the sauce entirely.\n- **A (8)** multiplies filling × sauce \\((4 \\times 2)\\) and drops the bread.\n- **D (14)** multiplies two categories and then *adds* the third \\((3 \\times 4 + 2)\\), mixing the\n operations.\n\nOnly **E (24)** multiplies all three.", + "hints": [ + "Each sandwich needs one bread AND one filling AND one sauce — \"and\" between independent choices means multiply.", + "Multiply the count at every stage; don't leave a category out." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0018", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "combinatorics", + "permutations", + "factorial", + "arrangements" + ], + "choices": { + "A": "15", + "B": "20", + "C": "25", + "D": "60", + "E": "120" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "In how many different ways can 5 distinct books be arranged in a row on a shelf?", + "explanation": "Arranging all 5 distinct books in order is a **permutation of 5 items**: the first position can be\nany of 5 books, the next any of the remaining 4, then 3, then 2, then 1:\n\n\\[5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120\\]\n\n**The traps:**\n\n- **A (15)** *adds* the descending counts \\((5 + 4 + 3 + 2 + 1)\\) instead of multiplying them.\n- **B (20)** stops after two positions \\((5 \\times 4)\\) — that counts arrangements of only 2 of the\n books, not all 5.\n- **D (60)** stops after three positions \\((5 \\times 4 \\times 3)\\), arranging only 3 books.\n- **C (25)** computes \\(5^2\\), as if each slot could independently be any of 5 books; but once a book\n is placed it can't be reused, so the pool shrinks each step.\n\nOnly **E (120)** multiplies all the way down to 1.", + "hints": [ + "Once a book is placed in a spot, it can't go in another — so the number of choices drops by one at each position.", + "Arranging all \\(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", + "combinations", + "committee", + "order-does-not-matter" + ], + "choices": { + "A": "21", + "B": "24", + "C": "56", + "D": "168", + "E": "336" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A committee of 3 people is to be selected from a group of 8 people. How many different committees\nare possible?", + "explanation": "A committee is a **selection where order does not matter** — the committee {A, B, C} is the same as\n{C, B, A}. That's a **combination**:\n\n\\[C(8,3) = \\frac{8 \\times 7 \\times 6}{3!} = \\frac{336}{6} = 56\\]\n\n**The traps:**\n\n- **E (336)** computes \\(8 \\times 7 \\times 6\\), a *permutation* — it counts every ordering of the same\n three people as different. Since order doesn't matter for a committee, you must divide by the\n \\(3! = 6\\) ways to arrange each trio.\n- **D (168)** starts from the 336 orderings but divides by 2 instead of by \\(3! = 6\\). Three chosen\n people can be arranged in 6 ways, not 2.\n- **A (21)** adds \\(8 + 7 + 6\\); **B (24)** multiplies \\(8 \\times 3\\). Neither reflects choosing 3\n distinct members from the group.\n\nOnly **C (56)** correctly divides out the ordering.", + "hints": [ + "Does rearranging the same three people give a *different* committee? If no, it's a combination, not a permutation.", + "Take the ordered count \\(8 \\times 7 \\times 6\\) and divide by \\(3!\\) to remove the duplicate arrangements." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0020", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "combinatorics", + "combinations", + "groups", + "and-rule" + ], + "choices": { + "A": "10", + "B": "16", + "C": "60", + "D": "126", + "E": "240" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A committee of 4 people is formed from 5 men and 4 women. How many committees consist of exactly\n2 men and 2 women?", + "explanation": "Build the committee in two independent stages — choose the men **and** choose the women — then\nmultiply the counts (order doesn't matter within either group, so use combinations):\n\n\\[C(5,2) \\times C(4,2) = 10 \\times 6 = 60\\]\n\nHere \\(C(5,2) = \\dfrac{5 \\times 4}{2} = 10\\) and \\(C(4,2) = \\dfrac{4 \\times 3}{2} = 6\\).\n\n**The traps:**\n\n- **B (16)** *adds* the two counts \\((10 + 6)\\) instead of multiplying. You need 2 men *and* 2 women,\n so the stages multiply.\n- **A (10)** stops after choosing the men \\((C(5,2))\\) and forgets to choose the women.\n- **D (126)** ignores the \"exactly 2 and 2\" requirement and just picks any 4 of the 9 people\n \\((C(9,4) = 126)\\).\n- **E (240)** treats the picks as ordered, using permutations \\(P(5,2) \\times P(4,2) = 20 \\times 12\\);\n but the two men on a committee aren't ranked, so order shouldn't be counted.\n\nOnly **C (60)** multiplies the two combinations.", + "hints": [ + "\"Exactly 2 men and 2 women\" is a choice from each group — choose men AND women, then multiply.", + "Within each group the members aren't ranked, so use combinations, and \\(C(5,2)=10\\), \\(C(4,2)=6\\)." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0021", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "permutations", + "arrangements", + "restriction", + "block-method" + ], + "choices": { + "A": "24", + "B": "48", + "C": "72", + "D": "120", + "E": "240" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "Five friends are seated in a row of 5 chairs. In how many of the possible seating arrangements\nare two particular friends, Ana and Ben, seated next to each other?", + "explanation": "Use the **block method**: since Ana and Ben must be adjacent, glue them into a single \"super-person.\"\nThat leaves **4 units** to arrange in a row:\n\n\\[4! = 24 \\text{ ways.}\\]\n\nBut inside the block, Ana and Ben can sit in either order (Ana–Ben or Ben–Ana), which doubles the\ncount:\n\n\\[4! \\times 2 = 24 \\times 2 = 48\\]\n\n**The traps:**\n\n- **A (24)** treats the pair as one block \\((4!)\\) but forgets that Ana and Ben can swap seats within\n the block — you must multiply by \\(2\\).\n- **D (120)** is \\(5!\\), the total number of arrangements with **no** restriction.\n- **C (72)** is the number of arrangements in which Ana and Ben are **not** together\n \\((120 - 48)\\) — the complement of what's asked.\n- **E (240)** doubles the full \\(5!\\) \\((2 \\times 120)\\), double-counting rather than restricting.\n\nOnly **B (48)** applies the block method correctly.", + "hints": [ + "Force the two friends together by treating them as a single block, then arrange the resulting units.", + "Don't forget the two people inside the block can still switch places with each other." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0022", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "combinations", + "at-least-one", + "complementary-counting" + ], + "choices": { + "A": "15", + "B": "60", + "C": "111", + "D": "126", + "E": "168" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A team of 4 is to be chosen from a group of 6 seniors and 3 juniors (9 people in all). How many\npossible teams include **at least one** junior?", + "explanation": "\"At least one\" invites **complementary counting**: count *all* teams, then subtract the teams that\nhave **no** junior (i.e., all seniors).\n\n- Total teams of 4 from 9: \\(C(9,4) = \\dfrac{9 \\times 8 \\times 7 \\times 6}{4!} = \\dfrac{3024}{24} = 126\\).\n- Teams with no junior (all 4 from the 6 seniors): \\(C(6,4) = C(6,2) = \\dfrac{6 \\times 5}{2} = 15\\).\n\n\\[\\text{At least one junior} = 126 - 15 = 111\\]\n\n**The traps:**\n\n- **D (126)** is the total number of teams — it forgets to remove the all-senior teams.\n- **A (15)** is the number of teams with **no** junior — the complement that should be *subtracted*,\n not reported.\n- **E (168)** comes from \"pick 1 junior to guarantee, then any 3 of the remaining 8\":\n \\(C(3,1) \\times C(8,3) = 3 \\times 56 = 168\\). This **over-counts** teams that have 2 or 3 juniors,\n because each such team gets counted once for each junior that could have been the \"guaranteed\" one.\n- **B (60)** counts only the teams with **exactly one** junior \\((C(3,1) \\times C(6,3) = 3 \\times 20)\\),\n missing teams with 2 or 3 juniors.\n\nOnly **C (111)** counts every team with at least one junior exactly once. (Check:\nexactly-1 \\(=60\\), exactly-2 \\(= C(3,2)C(6,2) = 3 \\times 15 = 45\\), exactly-3 \\(= C(3,3)C(6,1) = 6\\);\n\\(60 + 45 + 6 = 111\\).)", + "hints": [ + "\"At least one\" is usually fastest as total minus none — count all teams, then subtract the teams with zero juniors.", + "The \"zero juniors\" case means all 4 members come from the 6 seniors." + ] + }, { "id": "quant-counting-probability-probability-0007", "section": "quant", @@ -984,6 +1170,24 @@ "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 Arrangements & Selections", + "tags": [ + "combinatorics", + "counting-principle", + "permutations", + "combinations", + "factorial" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nCombinatorics is the art of counting *how many* ways something can happen without listing them\nall. On GMAT Focus Quant the numbers stay small, so the challenge is never the arithmetic — it is\ndeciding **which counting move applies**. Three questions sort out almost everything:\n\n1. Am I making a sequence of independent choices? → **multiply** (fundamental counting principle).\n2. Am I arranging items where **order matters**? → **permutation**.\n3. Am I selecting a group where **order does not matter**? → **combination**.\n\nGet the classification right and the formula is easy.\n\n## Core concepts\n\n**Fundamental counting principle.** If one stage can happen in \\(m\\) ways and a second, independent\nstage in \\(n\\) ways, the two together happen in \\(m \\times n\\) ways. This extends to any number of\nstages — you **multiply**, never add. (You add only when the choices are mutually exclusive\nalternatives — \"either this *or* that\" — not sequential steps.)\n\n**Factorial.** \\(n! = n \\times (n-1) \\times \\cdots \\times 2 \\times 1\\), the number of ways to arrange\n\\(n\\) distinct items in a row. By convention \\(0! = 1\\).\n\n**Permutations — order matters.** The number of ways to arrange \\(k\\) of \\(n\\) distinct items in\norder:\n\n\\[P(n,k) = \\frac{n!}{(n-k)!} = n \\times (n-1) \\times \\cdots \\times (n-k+1)\\]\n\nExample: 3 of 5 books in order = \\(5 \\times 4 \\times 3 = 60\\).\n\n**Combinations — order does not matter.** The number of ways to *select* \\(k\\) of \\(n\\) distinct\nitems when arrangement is irrelevant:\n\n\\[C(n,k) = \\binom{n}{k} = \\frac{n!}{k!\\,(n-k)!} = \\frac{P(n,k)}{k!}\\]\n\nA combination is just a permutation with the internal orderings divided out. Example: a committee of\n3 from 8 people = \\(\\dfrac{8 \\times 7 \\times 6}{3!} = \\dfrac{336}{6} = 56\\).\n\nTwo handy facts: \\(C(n,k) = C(n,\\,n-k)\\) (choosing who is *in* is the same as choosing who is *out*),\nand \\(C(n,0) = C(n,n) = 1\\).\n\n## Worked examples\n\n**Multiply the stages.** A meal is one appetizer (4 options), one entrée (5), and one dessert (3).\nMeals \\(= 4 \\times 5 \\times 3 = 60\\). Independent, sequential choices → multiply.\n\n**Arrange with a restriction.** In how many ways can 5 people sit in a row if two specific people\nmust sit together? Glue the pair into a single block: now 4 units arrange in \\(4! = 24\\) ways, and the\npair can switch places inside the block in \\(2! = 2\\) ways, giving \\(24 \\times 2 = 48\\).\n\n**Split into groups.** Choose a committee of 4 from 5 men and 4 women with exactly 2 of each. Choose\nthe men *and* the women, so multiply: \\(C(5,2) \\times C(4,2) = 10 \\times 6 = 60\\).\n\n**\"At least\" → count the complement.** From 9 people, how many teams of 4 include at least one of the\n3 juniors? Counting \"at least one\" directly means adding several cases; it is faster to take *all*\nteams minus the teams with *no* junior: \\(C(9,4) - C(6,4) = 126 - 15 = 111\\).\n\n## Common traps\n\n- **Adding when you should multiply.** Sequential independent stages multiply. Adding gives a number\n far too small.\n- **Permutation vs. combination.** If the answer to \"does rearranging the same items give a *different*\n outcome?\" is no, use a combination. A committee of {A, B, C} is the same committee as {C, B, A} — do\n not count the orderings. Using \\(P(n,k)\\) for a selection over-counts by a factor of \\(k!\\).\n- **Forgetting the internal order of a block.** When you glue items together for an\n \"adjacent\" restriction, remember to multiply by the arrangements *inside* the block.\n- **Attacking \"at least one\" head-on.** Complementary counting (total − none) is almost always cleaner\n and avoids double-counting the overlap cases.\n\n## Key takeaways\n\n- Independent, sequential choices **multiply**; mutually exclusive alternatives add.\n- **Order matters → permutation** \\(\\left(\\tfrac{n!}{(n-k)!}\\right)\\); **order doesn't → combination**\n \\(\\left(\\tfrac{n!}{k!(n-k)!}\\right)\\). A combination is a permutation divided by \\(k!\\).\n- Break \"and\" problems into stages and multiply the counts for each stage.\n- For \"at least one,\" count the total and subtract the \"none\" case." + }, { "id": "quant-statistics-descriptive", "section": "quant",