Recognise the difference between permutations and combinations and know when each should be used

Permutations and Combinations (Cambridge IGCSE 0606)

Learning Objective

Students will be able to:

• Distinguish between permutations (order matters) and combinations (order does not matter).
• Write and use the correct factorial formulas and notation.
• Decide quickly which method to apply in any exam question.
• Apply the formulas to everyday‑context problems while respecting the syllabus limits (no repetitions, no circular arrangements).

1. Factorial Notation

• Definition: For a non‑negative integer \(n\), \[ n! = n\times (n-1)\times (n-2)\times\cdots\times 2\times 1, \qquad 0! = 1. \]
• Factorials increase very rapidly – always cancel common factors before using a calculator.

2. Standard Notation

Concept	Standard Symbol(s)	Formula (factorials)
Permutation of \(r\) objects from \(n\)	\(^{n}P_{r}\) (or \(nPr\)) Alternative: P(n,r)	\[ ^{n}P_{r}=P(n,r)=\frac{n!}{(n-r)!} \]
Combination of \(r\) objects from \(n\)	\(^{n}C_{r}\) (or \(nCr\)) Alternative: C(n,r) or \(\displaystyle\binom{n}{r}\)	\[ ^{n}C_{r}=C(n,r)=\binom{n}{r}= \frac{n!}{r!\,(n-r)!} \]

3. Quick Decision Tree – “Which Formula?”

Is the problem about arranging objects? ──► Yes → Use Permutations (order matters)
│
No ──► Is it about selecting objects? ──► Yes → Use Combinations (order does NOT matter)
│
No ──► Check the wording – it may be outside the syllabus (e.g., circular arrangements,
      repetitions, mixed‑step problems).

In practice, ask yourself two questions:

1. Does the order in which the objects appear affect the answer?
2. Are repetitions allowed?

If the answer to (1) is “yes” and (2) is “no”, use the permutation formula; otherwise use the combination formula.

4. Syllabus Scope & What Is NOT Covered

• All objects are distinct and no repetitions are allowed.
• Circular arrangements (e.g., seating round a table) are excluded – a different formula is required and will not appear in the exam.
• Problems that combine a permutation step with a combination step in a single question are outside the required content; treat each step separately.
• Repetition‑allowed problems (e.g., “with replacement”) are not examined in this part of the syllabus.

Out‑of‑scope example: “In how many ways can 6 people be seated round a circular table?” – this will not be asked in the IGCSE 0606 paper.

5. Relationship Between the Two Formulas

Because a permutation is simply a combination with the \(r\) chosen objects arranged, the two are linked by

\[ ^{n}P_{r}=^{n}C_{r}\times r! \]

This identity is useful for checking work or converting between the two counts.

6. Worked Examples

Example 1 – Simple Permutation

Problem: A password consists of 4 different letters chosen from the 26 letters of the alphabet. How many possible passwords are there?

Order matters → permutations:

\[ ^{26}P_{4}= \frac{26!}{22!}=26\times25\times24\times23=358\,800. \]

There are 358 800 possible passwords.

Example 2 – Simple Combination

Problem: A student council needs to select a committee of 3 members from a class of 12 students. How many different committees can be formed?

Order does not matter → combinations:

\[ ^{12}C_{3}= \binom{12}{3}= \frac{12!}{3!\,9!}= \frac{12\times11\times10}{3\times2\times1}=220. \]

Thus, 220 distinct committees are possible.

Example 3 – Fixed Position (Pre‑condition)

Problem: Five different books are to be placed on a shelf. Book A must be at the left‑most position. In how many ways can the books be arranged?

Step 1 – Fix Book A at the left end (only 1 way).
Step 2 – Arrange the remaining 4 books: \(^{4}P_{4}=4!=24\).

Therefore, the total number of arrangements is \(1\times24=24\).

Example 4 – Tabular Counting Process (Combination)

Problem: From 5 different coloured pens (Red, Blue, Green, Yellow, Black) a student chooses 2 pens. List the counting process in a table and give the total number of selections.

First pen	Second pen (different)
Red	Blue, Green, Yellow, Black
Blue	Green, Yellow, Black
Green	Yellow, Black
Yellow	Black
Black	–

Counting the entries gives \(4+3+2+1=10\) selections, which matches the formula

\[ ^{5}C_{2}= \frac{5!}{2!\,3!}=10. \]

Example 5 – Algebraic Use of the Combination Formula

Problem: Express the number of ways to choose \(r\) objects from a set of \(2r+1\) distinct objects, and simplify.

\[ ^{\,2r+1}C_{r}= \frac{(2r+1)!}{r!\,(2r+1-r)!} = \frac{(2r+1)!}{r!\,(r+1)!}. \]

This is the simplest closed form for any integer \(r\ge0\).

7. Summary Table

Aspect	Permutation (\(^{n}P_{r}\))	Combination (\(^{n}C_{r}\))
Order	Important	Not important
Standard Symbol(s)	\(^{n}P_{r}\) (or \(nPr\)), P(n,r)	\(^{n}C_{r}\) (or \(nCr\)), C(n,r), \(\displaystyle\binom{n}{r}\)
Formula	\(\displaystyle\frac{n!}{(n-r)!}\)	\(\displaystyle\frac{n!}{r!\,(n-r)!}\)
Typical Wording	“In how many ways can … be arranged?”	“In how many ways can … be chosen?”
Example Context	Arranging books on a shelf, forming a password	Selecting a committee, forming a hand of cards

8. Common Mistakes to Avoid

• Confusing “arranged” with “selected”. Remember: order = permutation, no order = combination.
• Using the permutation formula for a combination problem – this over‑counts by a factor of \(r!\).
• Forgetting that \(0! = 1\). Consequently, \(^{n}P_{0}=1\) and \(^{n}C_{0}=1\).
• Applying the formulas when repetitions are allowed or when the arrangement is circular – these are outside the syllabus scope.
• Neglecting the relationship \(\displaystyle ^nP_r = ^nC_r \times r!\); it can be used to check answers.

9. Practice Questions

1. A committee of 4 is to be chosen from 9 people.
   Answer: \(^{9}C_{4}=126\).
2. How many 5‑digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without repetition?
   Answer: \(^{6}P_{5}=6\times5\times4\times3\times2=720\).
3. From a deck of 52 cards, how many ways can a hand of 3 cards be dealt if the order of dealing does not matter?
   Answer: \(^{52}C_{3}=22\,100\).
4. A school has 8 different clubs. In how many ways can a student join 2 clubs if the order of joining is irrelevant?
   Answer: \(^{8}C_{2}=28\).
5. Express the number of ways to arrange \(r\) distinct objects selected from a set of \(n=2r+3\) distinct objects, and simplify.
   Solution: \(^{\,2r+3}P_{r}= \dfrac{(2r+3)!}{(r+3)!}\).
6. Five books are to be placed on a shelf, but the thickest book must be at the right‑hand end. How many arrangements are possible?
   Solution: Fix the thickest book (1 way) and arrange the remaining 4 books: \(^{4}P_{4}=4!=24\).

10. Suggested Diagram (for hand‑out)