Ch 1 · 11 Permutations and Combinations · O Level Additional Mathematics 4037
Topic 1 of 3 in Series, Vectors and Counting — O Level Additional Mathematics 4037 — 58 lessons.
Introduction
How many 4-digit PINs use different digits? How many ways can a coach pick 5 players from a squad of 12? Counting rules answer these instantly. The key question is always: does the order matter? (Syllabus topic 11.)
Key ideas — Factorials
n! = n × (n - 1) × dots × 2 × 1, and 0! = 1. n different objects can be arranged in a line in n! ways: 7 different books on a shelf in 7! = 5040 ways.
🧑🏫 Teacher's explanation
Imagine you are lining up your friends for a photo. If you have 3 friends, you have 3 choices for the first spot. Once that person is chosen, you only have 2 choices left for the second spot, and finally 1 choice for the last spot. To find the total number of ways to arrange them, you multiply these choices together: 3 × 2 × 1 = 6. This 'countdown' multiplication is what we call a factorial.
📖 Definitions
🔍 Worked example — Easy
Calculate the value of 5!.
🔍 Worked example — Standard
A club has 6 members. How many different ways can they be arranged in a straight line for a group photograph?
🔍 Worked example — Exam-style
Simplify the expression 10!/8! without using a calculator.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Permutations: order matters
The number of ways to arrange r objects chosen from n different objects is
🧑🏫 Teacher's explanation
Imagine you are organizing a race. If you are just picking three runners to go get water, the order doesn't matter—it's just a group. But if you are handing out Gold, Silver, and Bronze medals, the order is everything! If Runner A gets Gold and Runner B gets Silver, that is a totally different outcome than Runner B getting Gold and Runner A getting Silver. This is the essence of a 'permutation'.
📖 Definitions
🔍 Worked example — Easy
Calculate the number of ways to arrange 4 different books on a shelf from a collection of 7 different books.
🔍 Worked example — Standard
A club has 10 members. How many ways can a President, a Secretary, and a Treasurer be chosen?
🔍 Worked example — Exam-style
In a race with 8 runners, how many different ways can the first 2 finishers be recorded?
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Combinations: order does not matter
The number of ways to select r objects from n different objects is
🧑🏫 Teacher's explanation
Imagine you are making a fruit salad. You pick an apple, a banana, and a cherry. If you pick the cherry first, then the apple, then the banana, you still have the exact same fruit salad. The order in which you grabbed them doesn't change the result. This is the essence of a 'combination'. In mathematics, we use combinations when we are 'selecting' or 'choosing' a group where the arrangement of the members doesn't matter.
📖 Definitions
🔍 Worked example — Easy
Calculate the number of ways to choose a committee of 3 students from a group of 8 students.
🔍 Worked example — Standard
A bag contains 10 different colored marbles. How many ways can you select 4 marbles at once?
🔍 Worked example — Exam-style
Prove that ^nC_r = ^nC_n-r using the algebraic definition.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Restrictions
In this syllabus, problems do not involve repeated objects, arrangements in a circle, or mixtures of permutations and combinations in one problem.
🧑🏫 Teacher's explanation
When we solve counting problems, we are essentially building a sequence of decisions. If a problem has a 'restriction' (a rule that limits your choices), you must deal with that rule first. Think of it like packing for a trip: if you know you MUST bring a specific heavy coat, you put that in the suitcase first, and then you see how much space is left for everything else. You don't try to pack the whole suitcase and then realize the coat won't fit!
📖 Definitions
🔍 Worked example — Easy
How many 3-digit numbers can be formed using the digits \1, 2, 3, 4, 5\ if no digit can be repeated and the number must be even?
🔍 Worked example — Standard
A committee of 5 people is to be chosen from 6 men and 4 women. How many ways can the committee be formed if it must contain at least 4 women?
🔍 Worked example — Exam-style
Find the number of ways to arrange the letters in the word 'MATHS' such that the letters 'M' and 'T' are never next to each other.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Algebraic problems
Write the formula out and simplify: ^nC_2 = n(n - 1)/2, ^nP_2 = n(n - 1).
🧑🏫 Teacher's explanation
Imagine you are organizing a small tournament. If you want to know how many ways you can award a Gold and a Silver medal to n players, the order matters because being 1st is different from being 2nd. This is a permutation. If you just want to pick 2 players to go buy snacks for the group, the order doesn't matter; picking Alice then Bob is the same as picking Bob then Alice. This is a combination.
📖 Definitions
🔍 Worked example — Easy
Simplify the expression for ^nP_2 in terms of n.
🔍 Worked example — Standard
Solve for n in the equation ^nC_2 = 15.
🔍 Worked example — Exam-style
Find the value of n if ^nP_2 = 3 imes ^nC_2.
⚠️ Common errors
eq$ BA), whereas combinations do not (e.g., AB = BA). Fix: Ask yourself: 'If I swap the order of the items I picked, does it result in a different outcome?' If yes, use $P$. If no, use $C$.
🧠 Quick check 1
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🎯 Exam question
Worked examples
How many 4-digit numbers can be made from 1, 2, 3, 4, 5, 6, 7 with no digit repeated? ^7P_4 = 840. How many are greater than 5000? The first digit is 5, 6 or 7: 3 × ^6P_3 = 360.
🔍 Worked example — Easy
How many 4-digit numbers can be formed using the digits \1, 2, 3, 4, 5, 6, 7\ if no digit can be repeated?
🔍 Worked example — Standard
A committee of 4 is to be chosen from 6 men and 5 women. How many ways can this be done if the committee must contain exactly 2 women?
🔍 Worked example — Exam-style
Given that ^nC_2 = 45, find the value of n.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 6! = 720. (2) ^7P_3 = 210. (3) ^8C_4 × ^4C_2 = 70 × 6 = 420. (4) n(n - 1) = 56 ⇒ n = 8. (5) 3 girls: ^6C_3 × ^7C_2 = 420; 4 girls: ^6C_4 × ^7C_1 = 105; 5 girls: ^6C_5 = 6; total 531.