← Senior Mathematics Studio

LEARN · EXPLAIN · REVISE

Use digits 0,…,t−1 to form two-digit numbers without repeated digits, where 2≤t≤9. Count them.

Read the idea, work independently, then explain what changed.

Return to the lesson / paper ↗

高三選擇性必修 第三册(A版).pdf · 6.1 · PDF 7 / printed page 2

TOPIC 01

Addition and multiplication counting principles

Count disjoint alternatives and sequential choices without overlap or omission.

What you will be able to explain

  • Count disjoint alternatives and sequential choices without overlap or omission.
  • Justify the method and check the conditions in a new situation.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Transfer#Worked example

Use digits 0,…,t−1 to form two-digit numbers without repeated digits, where 2≤t≤9. Count them.

t=6t=6
  • Addition requires disjoint cases; stage counts must hold for each branch or use separate branches.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Decide whether choices are alternative cases or successive stages, then check overlap.
Hint 2
Use this intermediate relation.
(t−1)(t−1)(t-1)(t-1)
Worked solution
  1. Decide whether choices are alternative cases or successive stages, then check overlap.

  2. Apply the stated relation and retain its conditions.

    first choices=5\text{first choices}=5
  3. Apply the stated relation and retain its conditions.

    second choices=5\text{second choices}=5
  4. Apply the stated relation and retain its conditions.

    N=25N=25
  5. The first digit cannot be zero; the second may be zero but cannot repeat the first.

The requested value is 25.

Checks and common pitfalls: The first digit cannot be zero; the second may be zero but cannot repeat the first.

Think first. Reveal a hint when the class is ready.

Focus on one question

Teacher preparation and assessment

Question sequence

  • Count disjoint alternatives and sequential choices without overlap or omission.
  • Which condition is essential in addition and multiplication counting principles?
  • When should two counts be added rather than multiplied?

Board plan

  • Defining relation: Count disjoint alternatives and sequential choices without overlap or omission.
    N=∑NiorN=∏NiN=\sum N_i\quad\text{or}\quad N=\prod N_i
  • Conditions: Addition requires disjoint cases; stage counts must hold for each branch or use separate branches.

Anticipated thinking

  • Having two choices to describe does not automatically mean multiply.

Assessment checklist

  • 1 mark: choose the correct representation and conditions.
  • 1 mark: establish the intermediate relation.
  • 1 mark: complete a connected calculation or proof.
  • 1 mark: interpret and check the conclusion.

No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.

Curriculum and source notes ↗