Year 10A - Counting principles and arrangements - Free practice

Year 10A Counting Principles, Factorials & Arrangements Worksheets & Practice

Master factorial notation and systematic counting, the foundation of senior probability.

Quick answer

How do you count arrangements using factorials and the multiplication principle?

The multiplication principle says: if one stage has m choices and the next has n choices, there are m × n total arrangements. Factorial notation gives the number of ways to order n distinct objects as n! = n × (n−1) × … × 1. For example, 4! = 24 ways to arrange 4 books.

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Worked examples

Year 10A counting principles and arrangements: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Evaluating and simplifying factorial expressions
Evaluate $\dfrac{8!}{6!}$ .
Show worked solution
Key idea: Cancel common factors by expanding just enough of the larger factorial to reach the smaller one.
1. Write $8!$ as $8 \times 7 \times 6!$ :
  $\frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!}$
2. Cancel $6!$ from the numerator and denominator:
  $= 8 \times 7$
3. Evaluate:
  $= 56$
Answer: $56$
Example 2 · Counting ordered arrangements of a subset
A school captain team consists of 10 candidates. In how many ways can a President, Vice-President and Treasurer be chosen from the 10 candidates, if no person can hold more than one role?
Show worked solution
Key idea: Because the three roles are different, order matters. Use the multiplication principle: count the choices at each stage and multiply.
1. Choose the President: 10 options.
2. Choose the Vice-President from the remaining 9 candidates: 9 options.
3. Choose the Treasurer from the remaining 8 candidates: 8 options.
4. Multiply the choices together:
  $10 \times 9 \times 8 = 720$
This can also be written using factorial notation as:
$\frac{10!}{(10-3)!} = \frac{10!}{7!} = 720$

Answer: 720 ways
Example 3 · Probability using counting principles
Eight athletes compete in a race. Assuming all finishing orders are equally likely, what is the probability that a particular athlete named Priya finishes first and a particular athlete named Marcus finishes second?
Show worked solution
Key idea: Use counting principles to find the number of favourable outcomes and the total number of outcomes, then form a probability.
1. Find the total number of ways the 8 athletes can finish (all ordered arrangements of 8 people):
  $8! = 40320$
2. Count the favourable outcomes: Priya is first and Marcus is second, so the remaining 6 athletes fill positions 3 to 8 in any order:
  $6! = 720$
3. Calculate the probability:
  $P = \frac{6!}{8!} = \frac{720}{40320}$
4. Simplify:
  $\frac{720}{40320} = \frac{1}{56}$
Answer: $\dfrac{1}{56}$

Common mistakes

Where students usually lose marks

Forgetting that 0! = 1

Student writes $0! = 0$ , so $\frac{5!}{5!} = \frac{120}{0}$ , which is undefined.

By definition, $0! = 1$ . This comes up often when all objects are arranged, since the formula $\frac{n!}{(n-n)!} = \frac{n!}{0!} = n!$ must still work. Always substitute $0! = 1$ before simplifying.

Expanding the full factorial instead of cancelling

For $\frac{10!}{7!}$ , the student fully expands both: $\frac{3628800}{5040} = 720$ .

Cancel shared factors first: write $10! = 10 \times 9 \times 8 \times 7!$ and cancel $7!$ immediately. This avoids large arithmetic and is much faster, especially in probability problems with big factorials.

Using multiplication when order does not matter (or vice versa)

A student counts ordered arrangements when the question asks only for a selection, multiplying unnecessarily.

Always ask: does the order of the outcome matter? If the question asks who gets which distinct role or which position, order matters and you multiply stage by stage. If the question asks only who is chosen (with no different roles), order does not matter and a different method is needed.

Off-by-one errors in restricted arrangement problems

Choosing 3 items from 7 in order, the student multiplies $7 \times 6 \times 6$ instead of $7 \times 6 \times 5$ .

Each stage reduces the available choices by exactly one, because one item has already been used. Write out the stages carefully: first choice has $n$ options, second has $n-1$ , third has $n-2$ , and so on.

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Your turn

Eight different runners finish a race in a random order. How many different arrangements of all eight runners are possible?

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About this worksheet

Why we built it

Counting principles and factorial notation sit at the heart of Year 10A probability and give students heading to senior Methods maths their first real toolkit for answering "how many ways?" questions systematically. These worksheets build from the multiplication principle through to permutations and factorial-based probability calculations, with worked solutions for every question so students can follow each step of the reasoning.

What's covered

Sub-skills your student will practise

✓Multiplication principle for multi-stage selections and arrangements
✓Factorial notation: evaluating n! and simplifying factorial expressions
✓Counting ordered arrangements (permutations) of n objects
✓Arrangements of subsets: selecting and ordering r objects from n
✓Counting arrangements with repeated or restricted elements
✓Calculating probabilities using counting principles and factorials

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Frequently asked

Questions parents ask about counting principles and arrangements

What is factorial notation and why does it matter in Year 10A?

Factorial notation writes the product of all positive integers up to n as n!, so 5! = 5 × 4 × 3 × 2 × 1 = 120. It gives a compact way to count arrangements, and it is used directly in probability calculations and in senior Mathematical Methods (Mathematics Advanced in NSW).

What is the difference between a permutation and a combination at this level?

A permutation counts ordered arrangements, so placing A then B is different from B then A. Year 10A focuses on permutations and the multiplication principle. Combinations, where order does not matter, are introduced more formally in senior maths.

How do restrictions or repeated elements change an arrangements problem?

If certain objects must stay together or certain positions are fixed, you reduce the count by treating those objects as a single unit first. Repeated identical objects reduce the total by dividing out the factorials of each repeated group.

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