Add combinatorics bundle to 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 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.

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, 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.

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, 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 \(n!\).

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, 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.

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, 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 \(C(5,2)=10\), \(C(4,2)=6\).

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, 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.