Handle fixed positions and leading-zero restrictions
Choose the most restricted position first and divide digit problems into disjoint endpoint cases when branch sizes differ.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.2. State whether order matters, objects are distinguishable, repetition is permitted and positions or rows are labelled. Count mutually exclusive cases before adding them; account for internal arrangements when using blocks. These lessons include seating in two or more rows. Arrangements in a circle are excluded from this syllabus section. Unless a question says otherwise, people and named roles are distinguishable, and selection is without replacement. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use the product rule without replacement.
- Recognise even digits and the place value of a leading zero.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which condition applies to a four-digit positive integer?
What you will learn
- Count arrangements with fixed or restricted positions.
- Separate integer leading-digit rules from code conventions.
A numeral and a code have different conventions
Using 0, 1, 2, 3 once each, there are 4! = 24 four-character codes. Four-digit positive integers cannot start with zero, so choose their first digit in three ways and arrange the remaining three digits in 3! ways: 18. Equivalently subtract the 3! = 6 codes that start with zero.
If a code is required to start with zero, zero is a fixed character, not an invalid leading digit. A four-character code from digits 0–5, with no repeated digit and initial zero, has 5 × 4 × 3 = 60 possibilities. Keep every position in the code; a leading zero is part of the recorded string.
Endpoint cases can have different branch sizes
For three-digit even integers from 0, 1, 2, 3, 4 without repetition, choose the final digit first. If it is 0, the leading digit has four choices and the middle digit three: 12 numbers. If it is 2 or 4, there are two final choices, three non-zero leading choices and three middle choices: 18 numbers. The total is 30.
A single product 3 × 4 × 3 is wrong here because the first-position count depends on whether zero was used at the end. Write disjoint cases and count each accurately. For fixed named objects in a row, remove their occupied positions before arranging the others; check that remaining restrictions still apply.
Pause and explain
Using 0, 1, 2, 3 once each, how many four-digit integers exist?
Worked example
Count four-digit integers using all of 0, 1, 2, 3 without repetition.
Show the worked solution
- The first digit is 1, 2 or 3, giving three choices.
- Arrange the three unused digits in the remaining three positions in 3! ways.
- The count is 3 × 6 = 18; zero is allowed in later positions.
Answer 18 integers.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Count four-digit integers using 0, 1, 2, 3 exactly once each.
Give me a hint
There are three allowed first digits, then unrestricted remaining positions.
Compare my reasoning
- Choose the leading digit in three non-zero ways.
- Arrange the remaining three digits in 3! = 6 ways.
- Multiply 3 × 6 = 18.
18 integers.
Look for these in your work
- Forbid zero only in the leading position.
- Use every digit exactly once.
Count three-digit even integers from 0, 1, 2, 3, 4 without repeated digits.
Give me a hint
Separate final digit zero from final digit two or four.
Compare my reasoning
- Final 0 gives 4 × 3 = 12 numbers.
- Final 2 or 4 gives 2 × 3 × 3 = 18 numbers, excluding zero from the leading position.
- These cases are disjoint, so total 12 + 18 = 30.
30 integers.
Look for these in your work
- Apply the leading-zero restriction in both cases.
- Explain why a single identical branch count would fail.
A four-character code uses digits 0–5 without repetition and must start with 0. Count codes and explain why the integer leading-zero rule does not exclude them.
Give me a hint
The first character is fixed; the other three are ordered unused digits.
Compare my reasoning
- Fix the initial zero in one way.
- Choose remaining characters in 5, 4 and 3 ways.
- The product is 60; a code records all four characters rather than integer digit length.
60 codes; their leading zero is a required character.
Look for these in your work
- Retain the stated code convention.
- Count the remaining ordered positions without replacement.
Common mistakes
- A leading-zero prohibition belongs to an integer, not automatically to a code.
- Choose restrictive positions before unrestricted ones.
- Split cases when available choices depend on earlier decisions.
What can you explain now?
Is 0123 a four-character code or a four-digit integer?
Compare with the explanation
It can be a four-character code, but its numerical value is the three-digit integer 123.
The string length and integer place-value convention are different.
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Make it stick
Start with restricted positions; distinguish a digit string from an integer and add disjoint endpoint cases.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.