Combinatorics path · C01

Before this lesson: Arithmetic and exact calculation

Your goal: Explain the empty arrangement and simplify only valid factorial expressions.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Factorials: the key idea

For a positive integer n, n!=1·2·⋯·n, and 0!=1. Factorials count ordered arrangements of distinct objects. The convention 0!=1 counts the single empty arrangement and makes identities such as n!=n(n−1)! valid at n=1. Cancel common products before calculating large factorials. A factorial ratio such as n!/(n−k)! counts k ordered choices without repetition when 0≤k≤n.

A worked example

Evaluate 10!/8! without computing 10!.

Expand only the unmatched factors: 10!=10·9·8!, so the ratio is 10·9=90.

Your turn: change one thing

How many orders can four different books occupy on a shelf?

Try this on paper before opening the explanation.

Compare your reasoning

There are 4 choices for the first position, then 3,2,1. The product is 4!=24.

Pause and check

A trap to avoid: Using factorials of negative integers or treating n! as n to the n.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Explain combinatorially why n!=n(n−1)! for n≥1.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Choose the first object in n ways. For each choice, the remaining n−1 objects have (n−1)! orders. The product rule gives n(n−1)!, and every full order has one unique first object and remaining arrangement.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. What is 0!?

Foundation

  1. 1
  2. 0
  3. Undefined
  4. −1
Hint

Count the empty arrangement.

Answer and reasoning

1. There is one way to arrange no objects.

2. What does n! count for n distinct objects?

Foundation

  1. Selections with repetition
  2. Linear orders of all objects
  3. Unordered subsets only
  4. Circular orders always
Hint

Order matters and every object is used once.

Answer and reasoning

Linear orders of all objects. The successive choices are n,n−1,…,1.

3. What is 5!?

Core

  1. 24
  2. 60
  3. 25
  4. 120
Hint

Multiply 5·4·3·2·1.

Answer and reasoning

120. The product is 120.

4. What is 10!/8!?

Core

  1. 80
  2. 2
  3. 45
  4. 90
Hint

Cancel 8!.

Answer and reasoning

90. Only 10·9 remains.

5. Why is 0!=1 useful in counting?

Stretch

  1. It counts the one empty arrangement
  2. It means zero objects are present twice
  3. It makes every factorial zero
  4. It is an approximation
Hint

An empty choice is still one possible outcome.

Answer and reasoning

It counts the one empty arrangement. The convention keeps product and factorial formulas consistent at boundary cases.

6. n!/(n−k)! counts what when 0≤k≤n?

Stretch

  1. Unordered selections always
  2. Selections allowing repetition
  3. Partitions of n
  4. Ordered selections of k distinct objects from n
Hint

Inspect the k descending factors.

Answer and reasoning

Ordered selections of k distinct objects from n. They count successive choices without replacement.

7. What is 1!?

Foundation

  1. 0
  2. 2
  3. Undefined
  4. 1
Hint

The product has one factor.

Answer and reasoning

1. 1!=1.

8. Evaluate 7!/5!.

Core

  1. 21
  2. 2
  3. 42
  4. 14
Hint

Cancel the common 5!.

Answer and reasoning

42. 7·6=42.

9. Why is n! not the count of circular orders with rotations identified?

Stretch

  1. Circular orders are always infinite
  2. There are no positions
  3. It distinguishes rotated copies
  4. Factorials never count arrangements
Hint

Linear orders choose an absolute first position.

Answer and reasoning

It distinguishes rotated copies. For distinct objects, each circular order has n rotated linear representations.

Choose your next step

Continue to Basic counting principles. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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