Know and use the notation n! and the standard expressions for permutations and combinations of n items taken r at a time
Permutations and Combinations – Cambridge IGCSE Additional Mathematics (Syllabus 11)
Learning Objectives
- Interpret and use factorial notation
n!. - Apply the standard formulas for:
- Permutations of n objects taken r at a time.
- Combinations of n objects taken r at a time.
- Distinguish clearly between situations where order matters (permutations) and where it does not (combinations).
- Connect these ideas to other parts of the IGCSE Additional Mathematics syllabus (e.g., probability, algebraic manipulation).
1. Factorial Notation
The factorial of a non‑negative integer n is the product of all positive integers up to n:
\[
n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1,\qquad 0! = 1
\]
| n | n! |
|---|---|
| 0 | 1 (by definition) |
| 1 | 1 |
| 2 | 2 |
| 3 | 6 |
| 4 | 24 |
| 5 | 120 |
Why factorials matter
- They give the total number of ways to arrange a set of distinct objects.
- They are the building blocks for permutation and combination formulas.
2. Permutations – Ordered Selections
A permutation is an arrangement of objects where the order of selection is important.
General Formula
\[
{}^{n}P_{r}= \frac{n!}{(n-r)!}\qquad\text{(read “n P r”) }
\]
Special Cases
- All objects selected (\(r=n\)): \({}^{n}P_{n}=n!\).
- One object selected (\(r=1\)): \({}^{n}P_{1}=n\).
- Choosing all but one (\(r=n-1\)): \({}^{n}P_{n-1}=n\,(n-1)!\).
Worked Example 1 – Arranging books
Four different books are to be placed on a shelf. How many possible arrangements?
\[
{}^{4}P_{4}= \frac{4!}{(4-4)!}= \frac{24}{1}=24
\]
Worked Example 2 – Line‑ups of students
From 8 students, how many different 3‑person line‑ups can be formed?
\[
{}^{8}P_{3}= \frac{8!}{5!}= \frac{8\times7\times6\times5!}{5!}=8\times7\times6=336
\]
Worked Example 3 – Using cancellation (large numbers)
How many ways can 5 officers be chosen from a club of 12 and then ordered (president, vice‑president, …)?
\[
{}^{12}P_{5}= \frac{12!}{7!}=12\times11\times10\times9\times8=95\,040
\]
3. Combinations – Unordered Selections
A combination is a selection of objects where the order does not matter.
General Formula
\[
{}^{n}C_{r}= \binom{n}{r}= \frac{n!}{r!\,(n-r)!}
\]
Special Cases
- \({}^{n}C{0}= {}^{n}C{n}=1\) – choosing none or all objects.
- \({}^{n}C{1}= {}^{n}C{n-1}=n\) – choosing a single object.
- \({}^{n}C_{2}= \dfrac{n(n-1)}{2}\) – useful for quick “pair” counts.
Worked Example 4 – Forming a committee
A committee of 3 is to be chosen from 10 people. How many possible committees?
\[
\binom{10}{3}= \frac{10!}{3!\,7!}= \frac{10\times9\times8}{3\times2\times1}=120
\]
Worked Example 5 – Selecting coloured balls
A box contains 6 different coloured balls. In how many ways can 2 balls be chosen when order does not matter?
\[
\binom{6}{2}= \frac{6!}{2!\,4!}= \frac{6\times5}{2}=15
\]
Worked Example 6 – Cancelling before calculating
How many ways can a team of 4 be chosen from 15 players?
\[
\binom{15}{4}= \frac{15!}{4!\,11!}= \frac{15\times14\times13\times12}{4\times3\times2\times1}=1365
\]
4. Relationship Between Permutations and Combinations
Every unordered selection of r objects can be ordered in r! different ways. Hence:
\[
{}^{n}P{r}= {}^{n}C{r}\times r! \qquad\text{and}\qquad
{}^{n}C{r}= \frac{{}^{n}P{r}}{r!}
\]
5. Quick‑Reference Summary Table
| Concept | Notation | Formula | When to use |
|---|---|---|---|
| Factorial | n! | \(n! = n\times(n-1)\times\cdots\times1,\;0! = 1\) | Fundamental building block |
| Permutation | \({}^{n}P_{r}\) or \(P(n,r)\) | \(\displaystyle {}^{n}P_{r}= \frac{n!}{(n-r)!}\) | Order matters (arrangements, line‑ups, passwords) |
| Combination | \({}^{n}C_{r}\) or \(\displaystyle\binom{n}{r}\) | \(\displaystyle \binom{n}{r}= \frac{n!}{r!\,(n-r)!}\) | Order does not matter (committees, selections, hand‑cards) |
6. Common Pitfalls & How to Avoid Them
- Mis‑identifying order: Look for words such as “arrange”, “line‑up”, “password” (permutations) versus “choose”, “select”, “form a committee” (combinations).
- Missing the \((n-r)!\) term in permutations: Always write the full formula before simplifying.
- Forgetting to divide by \(r!\) for combinations: Start from the permutation count and divide by \(r!\) if you’re unsure.
- Zero factorial confusion: Remember \(0! = 1\); it appears in edge cases like \(\binom{n}{0}\) or \(\binom{n}{n}\).
- Handling large numbers: Cancel common factorial factors early. Example:
\[
{}^{8}P_{3}= \frac{8!}{5!}=8\times7\times6=336
\]
rather than computing full 8! and 5!.
- Mixing up notation: \({}^{n}P{r}\) and \({}^{n}C{r}\) are distinct; keep a separate notebook page for each.
7. Practice Questions (with marks)
- Passwords (4 marks)
How many different 5‑letter passwords can be formed from the letters A, B, C, D, E, F if no letter may be repeated?
Solution: \(^{6}P_{5}= \dfrac{6!}{1!}=6\times5\times4\times3\times2=720\).
- Card hands (5 marks)
From a standard deck of 52 playing cards, in how many ways can a hand of 4 cards be chosen?
Solution: \(\displaystyle\binom{52}{4}= \frac{52\times51\times50\times49}{4\times3\times2\times1}=270\,725\).
- Club selection (3 marks)
A school offers 12 different clubs. In how many ways can a student join exactly 3 clubs?
Solution: \(\displaystyle\binom{12}{3}= \frac{12\times11\times10}{3\times2\times1}=220\).
- Word arrangements (2 marks)
In how many distinct ways can the letters of the word “MATHS” be arranged?
Solution: All letters are different, so \(5! = 120\).
- Ordered relay team (4 marks)
Two runners are to be selected from a group of 8 to run a relay race. In how many ways can the two runners be ordered (first and second leg)?
Solution: \(^{8}P_{2}=8\times7=56\).
- Committee vs. line‑up (6 marks)
A school council has 9 members.
- How many 4‑member sub‑committees can be formed?
- How many ways can the same 4 members be arranged as a line‑up for a photo?
Solution: (a) \(\displaystyle\binom{9}{4}=126\). (b) Multiply by \(4!\): \(126\times24=3\,024\).
8. Visual Aid – Tree Diagram
9. How This Topic Connects to the Rest of the Syllabus
Permutations and combinations underpin many other IGCSE Additional Mathematics sections:
- Probability (Section 12) – The probability of an event is \(\dfrac{\text{favourable outcomes}}{\text{total outcomes}}\); counting outcomes often uses permutations/combinations.
- Algebraic manipulation (Sections 3‑5) – Simplifying factorial expressions reinforces factorisation skills.
- Exponential & logarithmic equations (Section 6) – Large factorials can be approximated using Stirling’s formula, linking to growth concepts.
- Coordinate geometry (Section 8) – Counting lattice points or arranging points on a circle may involve combinatorial reasoning.
Mastering factorial notation and the two core formulas therefore provides a solid foundation for the remainder of the IGCSE Additional Mathematics course.