Combinations — Selecting Items Without Order

🎯 Learning Objectives

• Calculate the number of combinations of n items taken r at a time
• Understand the binomial coefficient ${n \choose r}$ and its properties
• Distinguish between permutations and combinations
• Apply combinations to probability problems
• Use Pascal's triangle to find binomial coefficients

📋 Prerequisites

• Permutations (09-permutations) — understanding when order matters
• Counting Fundamentals (08-counting-fundamentals) — factorial notation
• Basic algebra

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📖 Core Content

11.1 Intuition: When Order Doesn't Matter

• Permutation: Pick 3 people for 1st, 2nd, 3rd prizes → ABC ≠ ACB → order matters
• Combination: Pick 3 people for a committee → {A,B,C} = {C,A,B} → order doesn't matter

Everyday analogy: Choosing 3 toppings for a pizza: {mushrooms, pepperoni, olives} is the same pizza no matter what order you name the toppings. That's a combination. 🔑 Key Insight: Every permutation is a combination plus an ordering. So:

$P(n, r) = C(n, r) \times r!$

To go from combinations to permutations, multiply by $r!$ (the ways to order the r selected items).

11.2 The Combination Formula

11.2.1 Derivation

$$C(n,r) = \frac{P(n,r)}{r!} = \frac{n!}{(n-r)! \times r!}$$

11.2.2 Notation

$$C(n, r) = {n \choose r} = \frac{n!}{r!(n-r)!}$$

11.2.3 Example

$$C(10, 3) = {10 \choose 3} = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$$

11.3 Properties of Combinations

Symmetry

$${n \choose r} = {n \choose n-r}$$

Special Values

Recurrence Relation (Pascal's Rule)

$${n \choose r} = {n-1 \choose r} + {n-1 \choose r-1}$$

11.4 Pascal's Triangle

pseudo
          1                     ${0 \choose 0}$
        1   1                   ${1 \choose 0}$ ${1 \choose 1}$
      1   2   1                 ${2 \choose 0}$ ${2 \choose 1}$ ${2 \choose 2}$
    1   3   3   1               ${3 \choose 0}$ ${3 \choose 1}$ ${3 \choose 2}$ ${3 \choose 3}$
  1   4   6   4   1             ${4 \choose 0}$ ${4 \choose 1}$ ${4 \choose 2}$ ${4 \choose 3}$ ${4 \choose 4}$
1   5  10  10   5   1           etc.

1. Find binomial coefficients quickly
2. Expand binomials $(a + b)^n$ (the $n^{th}$ row gives coefficients)
3. Solve combination problems without formulas

11.5 Combinations with Repetition

$${n + r - 1 \choose r} = {n + r - 1 \choose n - 1}$$ $${5 + 3 - 1 \choose 3} = {7 \choose 3} = 35$$

Note: This is more advanced and less commonly tested in BSMA1002. Focus on combinations without repetition first.

11.6 Decision Flowchart

11.7 Worked Examples

Example 1: Basic Combination (Easy)

$$C(8, 3) = {8 \choose 3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$$

Example 2: Probability Using Combinations (Medium)

$$C(9, 3) = 84$$ $$C(5, 2) \times C(4, 1) = 10 \times 4 = 40$$ $$P(2R, 1B) = \frac{40}{84} = \frac{10}{21} \approx 0.476$$

Example 3: Committee Problems (Harder)

$$C(10, 5) = 252$$ $$C(6, 3) \times C(4, 2) = 20 \times 6 = 120$$

• 3 women, 2 men: $C(4, 3) \times C(6, 2) = 4 \times 15 = 60$
• 4 women, 1 man: $C(4, 4) \times C(6, 1) = 1 \times 6 = 6$ Total with at least 3 women: $60 + 6 = 66$

11.8 Edge Cases & Gotchas

When r > n

When r is Negative

The Sum of Combinations

$${n \choose 0} + {n \choose 1} + \cdots + {n \choose n} = 2^n$$

11.9 Why This Matters

• Binomial distribution (Week 11): $P(X = k) = {n \choose k} p^k (1-p)^{n-k}$
• Hypergeometric distribution: Sampling without replacement
• Machine learning: Feature selection (which subset of features?)
• Lottery design: How many possible lottery tickets?

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📐 Key Formulas / Concepts

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⚠️ Common Pitfalls

Pitfall 1: Confusing Permutations and Combinations

• Permutation = Position matters
• Combination = Choose (order doesn't matter)

Pitfall 2: Forgetting to Multiply Cases

Pitfall 3: Using Combinations When Items Are Distinct but Repetition Is Allowed

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📝 Practice Questions

Q1: Basic Combination

</strong>

Compute ${8 \choose 3}$.

<details> <strong>Solution</strong>

${8 \choose 3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

$\boxed{56}$

</details> > **Q2: Committee Formation** > > </strong> > > How many ways to choose a committee of 4 from 15 people? > > <details> <strong>Solution</strong> > > $C(15, 4) = \frac{15!}{4!11!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = \frac{32760}{24} = 1365$ > > $\boxed{1,365}$ </details> > **Q3: Compare P and C** > > </strong> > > Compute both $P(8,3)$ and $C(8,3)$. How are they related? > > <details> <strong>Solution</strong> > > $P(8,3) = 8 \times 7 \times 6 = 336$ $C(8,3) = \frac{8!}{3!5!} = 56$ > > Relationship: $P(8,3) = C(8,3) \times 3! = 56 \times 6 = 336$ ✅ > > Each combination of 3 items can be ordered in 3! = 6 ways. </details> > **Q4: Lottery Probability** > > </strong> > > In a lottery, you choose 6 numbers from 1-49. How many possible tickets? > > <details> <strong>Solution</strong> > > $C(49, 6) = \frac{49!}{6!43!} = \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{10,068,347,520}{720} = 13,983,816$ > > So there's a 1 in 13,983,816 chance of winning with one ticket. > > $\boxed{13,983,816}$ </details> > **Q5: Probability with Combinations** > > </strong> > > A bag has 7 red and 3 blue marbles. Draw 4 without replacement. Probability of exactly 2 red? > > <details> <strong>Solution</strong> > > **Total ways:** $C(10, 4) = 210$ > > **Ways to get 2 red and 2 blue:** $C(7, 2) \times C(3, 2) = 21 \times 3 = 63$ > > **Probability:** $63/210 = 3/10 = 0.3$ > > $\boxed{0.3}$ </details> > **Q6: Symmetry Property** > > </strong> > > Verify: ${7 \choose 4} = {7 \choose 3}$ > > <details> <strong>Solution</strong> > > ${7 \choose 4} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = \frac{840}{24} = 35$ > > ${7 \choose 3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35$ > > ✅ They're equal. Choosing 4 items leaves behind 3 items. </details> > **Q7: "At Least" Problem** > > </strong> > > A group of 12 people — 7 men and 5 women. Choose a committee of 5. How many ways to get at least 4 men? > > <details> <strong>Solution</strong> > > **Case 1: 4 men, 1 woman** $C(7, 4) \times C(5, 1) = 35 \times 5 = 175$ > > **Case 2: 5 men, 0 women** $C(7, 5) \times C(5, 0) = 21 \times 1 = 21$ > > **Total:** $175 + 21 = 196$ > > $\boxed{196}$ </details> > **Q8: Poker Hand** > > </strong> > > A standard deck has 52 cards. How many 5-card poker hands are possible? How many are "flushes" (all 5 same suit)? > > <details> <strong>Solution</strong> > > **Total hands:** $C(52, 5) = 2,598,960$ > > **Flushes:** Choose 1 suit (4 choices), then choose 5 cards from that suit (13 cards): $4 \times C(13, 5) = 4 \times 1,287 = 5,148$ > > $\boxed{2,598,960 \text{ total},\ 5,148 \text{ flushes}}$ </details> > **Q9: Application — Quality Control** > > </strong> > > A batch of 20 items has 4 defective. You inspect 5 at random. Probability of finding exactly 1 defective? > > <details> <strong>Solution</strong> > > **Total ways to inspect 5 from 20:** $C(20, 5)$ > > **Ways to get 1 defective (from 4) and 4 good (from 16):** $C(4,1) \times C(16,4)$ > > $P = \frac{C(4,1) \times C(16,4)}{C(20,5)} = \frac{4 \times 1820}{15504} = \frac{7280}{15504} \approx 0.4695$ > > $\boxed{P \approx 0.47}$ </details> > **Q10: Prove Pascal's Rule** > > </strong> > > Verify Pascal's rule for n = 6, r = 3: ${6 \choose 3} = {5 \choose 3} + {5 \choose 2}$ > > <details> <strong>Solution</strong> > > ${6 \choose 3} = \frac{6!}{3!3!} = \frac{720}{36} = 20$ > > ${5 \choose 3} = \frac{5!}{3!2!} = \frac{120}{12} = 10$ > > ${5 \choose 2} = \frac{5!}{2!3!} = \frac{120}{12} = 10$ > > $10 + 10 = 20 = {6 \choose 3}$ ✅ > > The rule holds: each Pascal's triangle entry is the sum of the two above it. </details> * * * ## 🔗 Cross-References - **Next topic:** [Probability — Basic Concepts](/notes/01-foundation-bsma1002-stats-1-week07-11-probability-intro) — applying counting to probability calculations - **Previous:** [Permutations](/notes/01-foundation-bsma1002-stats-1-week06-09-permutations) — understanding the permutation-combination relationship - **Week 11 (Bernoulli & Binomial):** The binomial coefficient in $P(X=k)$ - **BSMA1001-maths-1:** The binomial theorem $(a+b)^n$ - **BSMA1004 (Stats 2):** Combinations in sampling distributions [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Permutations**](/notes/01-foundation-bsma1002-stats-1-week06-09-permutations)[Next**Probability Basics**](/notes/01-foundation-bsma1002-stats-1-week07-11-probability-intro)