LEARN · EXPLAIN · REVISE

Passwords of length 3 use t symbols with repetition allowed. Exclude those with all symbols identical. Count remaining passwords.

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

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.

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

01 / Standard#Your turn

t=11

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.

N=t^3-t

Worked solution

Decide whether choices are alternative cases or successive stages, then check overlap.
Apply the stated relation and retain its conditions.
$N=1331-11=1320$
The excluded all-identical passwords are one per symbol, not one total.

The requested value is 1320.

Checks and common pitfalls: The excluded all-identical passwords are one per symbol, not one total.

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

Teacher preparation and assessment

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?

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