How to Solve: Permutations Basics

How to Solve: Permutations Basics

Introduction

"Imagine you’re organizing a school photo with 5 friends—how many different ways can you line up? Permutations give you the exact answer in seconds, and they’re a guaranteed question on your next exam!

What You Need To Know First

Before diving into permutations, ensure you understand: 1. Factorials – The product of all positive integers up to a number (e.g., 4! = 4 × 3 × 2 × 1 = 24). 2. Order Matters – In permutations, ABC is different from BAC. 3. Basic Counting Principle – If one event can happen m ways and another n ways, together they happen m × n ways.

Key Vocabulary

Formulas To Know

1. Permutation Formula (Order Matters, No Repetition)

Formula: nPr = n! / (n - r)!

Variables: - n = Total number of distinct items. - r = Number of items being arranged. - ! = Factorial (e.g., 4! = 24).

Memorise This? ✅ YES (Not always given on exam sheets.)

When to Use: - Arranging r items from n distinct items where order matters (e.g., race positions, passwords, seating).

2. Permutation with Repetition (Items Can Repeat)

Formula: n^r

Variables: - n = Number of choices for each position. - r = Number of positions to fill.

Memorise This? ✅ YES (Often given, but know when to use it.)

When to Use: - When items can be repeated (e.g., 3-digit PINs where digits can repeat: 10^3 = 1000 possible PINs).

3. Permutation of All Items (Full Arrangement)

Formula: n!

Variables: - n = Total number of distinct items.

Memorise This? ✅ YES (Same as nPn.)

When to Use: - Arranging all items (e.g., 5 books on a shelf: 5! = 120 ways).

Step-by-Step Method

How to Solve Any Permutation Problem

Step 1: Read the Question Carefully - Underline key numbers and words like "arrange," "order," "line up," "first/second/third." - Ask: Does order matter? If yes → permutation.

Step 2: Identify n and r - n = Total number of items to choose from. - r = Number of items being arranged.

Step 3: Check for Repetition - Can items repeat? (e.g., digits in a phone number, letters in a password.) - Yes → Use n^r. - No → Use nPr = n! / (n - r)!.

Step 4: Plug into the Correct Formula - Write the formula clearly. - Substitute n and r. - Simplify step-by-step (show all working).

Step 5: Calculate Factorials - Break down factorials to avoid mistakes (e.g., 5! = 5 × 4 × 3 × 2 × 1). - Cancel terms where possible (e.g., 5! / 3! = (5 × 4 × 3!) / 3! = 5 × 4 = 20).

Step 6: Write the Final Answer - Include units if needed (e.g., "ways," "arrangements"). - Box or underline the answer.

Worked Example Using the Steps

Question: How many different ways can 4 students (A, B, C, D) be arranged in a line for a photo if only 2 are chosen at a time?

Step 1: Order matters (ABC is different from BAC). → Permutation. Step 2: n = 4 (total students), r = 2 (chosen at a time). Step 3: No repetition (each student is unique). Step 4: Use nPr = n! / (n - r)!. Step 5: 4P2 = 4! / (4 - 2)! = 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 12. Step 6: Answer: 12 ways.

Worked Examples

Example 1 – Basic (No Tricks)

Question: How many different 3-letter passwords can be made from the letters {A, B, C, D} if no letter is repeated?

Solution: 1. Order matters (ABC ≠ BAC) → Permutation. 2. n = 4 (letters), r = 3 (letters in password). 3. No repetition → Use nPr. 4. 4P3 = 4! / (4 - 3)! = 4! / 1! = 24 / 1 = 24. 5. Answer: 24 possible passwords.

What We Did and Why: - We used nPr because order matters and letters can’t repeat. - 4P3 means we’re arranging 3 out of 4 letters.

Example 2 – Medium (Added Complication)

Question: A race has 8 runners. How many different ways can gold, silver, and bronze medals be awarded?

Solution: 1. Order matters (gold ≠ silver ≠ bronze) → Permutation. 2. n = 8 (runners), r = 3 (medals). 3. No repetition (one runner can’t win multiple medals) → Use nPr. 4. 8P3 = 8! / (8 - 3)! = 8! / 5! = (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336. 5. Answer: 336 possible ways.

What We Did and Why: - We canceled 5! to simplify the calculation. - 8P3 gives the number of ways to arrange 3 medals out of 8 runners.

Example 3 – Exam Style (Disguised, Time Pressure)

Question: A school committee has 6 members. They need to elect a president, vice-president, and secretary. How many different ways can these 3 positions be filled?

Solution: 1. Order matters (president ≠ vice-president) → Permutation. 2. n = 6 (members), r = 3 (positions). 3. No repetition (one person can’t hold multiple positions) → Use nPr. 4. 6P3 = 6! / (6 - 3)! = 6! / 3! = (6 × 5 × 4 × 3!) / 3! = 6 × 5 × 4 = 120. 5. Answer: 120 ways.

What We Did and Why: - The question is disguised as an election, but it’s just a permutation problem. - 6P3 calculates the number of ways to assign 3 distinct roles to 6 people.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in for your exam. Permutations are all about order mattering—like arranging friends in a photo or assigning medals in a race. Here’s the game plan: 1. Spot the keywords: ‘Arrange,’ ‘order,’ ‘line up,’ ‘first/second/third.’ If you see these, think permutations. 2. Find n and r: n is the total number of items, r is how many you’re arranging. 3. Check for repetition: If items can repeat (like digits in a PIN), use n^r. If not, use nPr = n! / (n - r)!. 4. Simplify early: Cancel factorials before multiplying to save time. 5. Double-check: Did you use the right formula? Did you subtract r from n in the denominator?

That’s it. Permutations are just factorials with a twist. Practice a few problems tonight, and you’ll own this on exam day. You’ve got this!