Permutations — Arranging Items in Order

🎯 Learning Objectives

• Calculate the number of permutations of n distinct items taken r at a time
• Distinguish between permutations with and without repetition
• Recognize when order matters in a counting problem
• Solve problems involving arrangements, seating, and ranking
• Understand circular permutations and when to use them

📋 Prerequisites

• Counting Fundamentals (08-counting-fundamentals) — multiplication rule and factorial notation
• Basic algebra: simplifying fractions

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📖 Core Content

10.1 Intuition: When Order Matters

• Arranging 3 books on a shelf: ABC ≠ ACB → Order matters → Permutation
• Choosing 3 students for a committee: {A,B,C} = {C,B,A} → Order doesn't matter → Combination (next topic)

Everyday analogy: A combination lock is actually a permutation lock! The order of numbers matters: 1-2-3 is different from 3-2-1. Real combinations (order doesn't matter) would make locks very insecure. 🔑 Key Insight: If the problem uses words like "arrange," "order," "rank," "line up," "sequence," or "first, second, third," it's likely a permutation.

10.2 Permutations of n Distinct Items Taken r at a Time

10.2.1 Intuition

• First prize: 10 choices
• Second prize: 9 remaining choices
• Third prize: 8 remaining choices

$$10 \times 9 \times 8 = 720$$

10.2.2 Formula

$$P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)$$

• $n$ = total number of items
• $r$ = number of items to arrange ($r \leq n$)
• $P(n, r)$ = number of permutations

10.2.3 Verification with Example

$$P(10, 3) = \frac{10!}{(10-3)!} = \frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720$$

10.3 Permutations of All n Items

$$P(n, n) = n!$$

10.4 Permutations with Repetition

$$n^r$$

10.5 Permutations with Identical Items

$$\frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}$$

Example

$$\frac{11!}{1! \times 4! \times 4! \times 2!} = \frac{39,916,800}{1 \times 24 \times 24 \times 2} = \frac{39,916,800}{1,152} = 34,650$$

10.6 Circular Permutations

$$(n-1)!$$

Example

$$(6-1)! = 5! = 120$$

10.7 Decision Flowchart

10.8 Worked Examples

Example 1: Basic Permutation (Easy)

$$P(8, 3) = 8 \times 7 \times 6 = 336$$

Example 2: Arranging Letters (Medium)

$$\frac{6!}{1! \times 3! \times 2!} = \frac{720}{1 \times 6 \times 2} = \frac{720}{12} = 60$$

Example 3: Arrangements with Restrictions (Harder)

• Treat the 4 women as a block: now we have 5 men + 1 block = 6 "items"
• Arrange the 6 items: 6! = 720
• Arrange women within the block: 4! = 24
• Total: $720 \times 24 = 17,280$ c) Men and women alternate:
• Since there are 5 men and 4 women, the row must start and end with men: M W M W M W M W M
• Arrange men in their 5 positions: 5! = 120
• Arrange women in their 4 positions: 4! = 24
• Total: $120 \times 24 = 2,880$

10.9 Edge Cases & Gotchas

When r > n

The Difference Between "Distinct" and "Identical"

Zero Factorial

10.10 Why This Matters

• Probability: Calculating probabilities of specific arrangements
• Cryptography: Password strength analysis
• Genetics: DNA sequence analysis
• Scheduling: Creating tournament brackets, seating charts
• Statistics: The number of possible rankings in non-parametric tests

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📐 Key Formulas / Concepts

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⚠️ Common Pitfalls

Pitfall 1: Confusing Permutations and Combinations

Pitfall 2: Forgetting to Divide by Identical Item Factorials

Pitfall 3: Confusing Linear and Circular Permutations

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📝 Practice Questions

Q1: Basic Permutation

How many ways can 5 people be arranged in a line?

<details> <strong>Solution</strong>

$P(5,5) = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$\boxed{120}$

</details> > **Q2: Partial Permutation** > > A class has 12 students. How many ways can a president, vice-president, and secretary be chosen? > > <details> <strong>Solution</strong> > > $P(12,3) = 12 \times 11 \times 10 = 1,320$ > > $\boxed{1,320}$ </details> > **Q3: With Identical Items** > > </strong> > > How many distinct words can be formed from "BOOKKEEPER"? > > <details> <strong>Solution</strong> > > Letters: B(1), O(2), K(2), E(2), P(1), R(1) → total 10 letters. > > $\frac{10!}{1! \times 2! \times 2! \times 2! \times 1! \times 1!} = \frac{3,628,800}{2 \times 2 \times 2} = \frac{3,628,800}{8} = 453,600$ > > $\boxed{453,600}$ </details> > **Q4: Circular Arrangement** > > </strong> > > In how many ways can 7 friends sit around a campfire (circular arrangement)? > > <details> <strong>Solution</strong> > > $(7-1)! = 6! = 720$ > > If the campfire had designated seating (positions marked), it would be $7! = 5,040$. > > $\boxed{720}$ </details> > **Q5: With Repetition** > > </strong> > > How many 3-letter sequences can be formed from the 26 letters if repetition is allowed? > > <details> <strong>Solution</strong> > > $26^3 = 17,576$ > > $\boxed{17,576}$ </details> > **Q6: Restrictions** > > </strong> > > How many ways can 4 men and 3 women be seated in a row if the men must sit together? > > <details> <strong>Solution</strong> > > Treat the 4 men as one block: now we have 1 block + 3 women = 4 items. > > Arrange 4 items: $4! = 24$ Arrange men within block: $4! = 24$ > > Total: $24 \times 24 = 576$ > > $\boxed{576}$ </details> > **Q7: Evaluate P(12,4)** > > </strong> > > Compute $P(12,4)$ using the formula. > > <details> <strong>Solution</strong> > > $P(12,4) = \frac{12!}{(12-4)!} = \frac{12!}{8!} = 12 \times 11 \times 10 \times 9 = 11,880$ > > $\boxed{11,880}$ </details> > **Q8: True or False** > > </strong> > > Determine if each statement is true or false: > > a) $P(5,3) = P(5,2)$ b) $P(7,7) = 7!$ c) $P(10,1) = 10$ > > <details> <strong>Solution</strong> > > a) **False.** $P(5,3) = 5 \times 4 \times 3 = 60$, while $P(5,2) = 5 \times 4 = 20$. They're different. > > b) **True.** Arranging all 7 items: $P(7,7) = 7! = 5,040$. > > c) **True.** $P(10,1) = 10$. There are 10 ways to choose 1 item from 10. </details> > **Q9: Application** > > </strong> > > A license plate has 2 letters (A-Z) followed by 4 digits (0-9). Letters and digits can repeat. How many possible plates? If letters cannot repeat, how many? > > <details> <strong>Solution</strong> > > **With repetition allowed:** $26^2 \times 10^4 = 676 \times 10,000 = 6,760,000$ > > **Without repetition (letters):** $P(26,2) \times 10^4 = 26 \times 25 \times 10,000 = 650 \times 10,000 = 6,500,000$ > > $\boxed{6,760,000 \text{ with repetition},\ 6,500,000 \text{ without}}$ </details> > **Q10: Advanced Restriction** > > </strong> > > How many ways can the letters of "ALGEBRA" be arranged if the two As must NOT be together? > > <details> <strong>Solution</strong> > > **Step 1:** Total arrangements of "ALGEBRA": Letters: A(2), L(1), G(1), E(1), B(1), R(1) → 7 letters $\frac{7!}{2!} = \frac{5040}{2} = 2520$ > > **Step 2:** Arrangements where As are together: Treat AA as one block: now we have 6 items (block + L, G, E, B, R) $6! = 720$ (No need to divide by 2! for the As since they're a block and identical) > > **Step 3:** Arrangements where As are NOT together: $2520 - 720 = 1800$ > > $\boxed{1,800}$ </details> * * * ## 🔗 Cross-References - **Next topic:** [Combinations](/notes/01-foundation-bsma1002-stats-1-week06-10-combinations) — selecting items when order doesn't matter - **Previous:** [Counting Fundamentals](/notes/01-foundation-bsma1002-stats-1-week05-08-counting-fundamentals) — the multiplication rule - **Week 7 (Probability):** Using permutations to calculate probabilities - **Week 11 (Binomial Distribution):** The binomial coefficient - **BSMA1001-maths-1:** Factorial notation from mathematics [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Counting Fundamentals**](/notes/01-foundation-bsma1002-stats-1-week05-08-counting-fundamentals)[Next**Combinations**](/notes/01-foundation-bsma1002-stats-1-week06-10-combinations)