Add combinatorics bundle: lesson + 7 questions (quant/counting-probability)

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

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+---
+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.

content/questions/quant/quant-counting-probability-combinatorics-0017.md

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+---
+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.

content/questions/quant/quant-counting-probability-combinatorics-0018.md

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+---
+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 \(n!\).

content/questions/quant/quant-counting-probability-combinatorics-0019.md

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+---
+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.

content/questions/quant/quant-counting-probability-combinatorics-0020.md

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+---
+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.

content/questions/quant/quant-counting-probability-combinatorics-0021.md

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+---
+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.
+