LEARN · EXPLAIN · REVISE

Count arrangements containing r identical As, two identical Bs and one C.

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

高三選擇性必修第三册(A版).pdf · 6.2 · PDF 19 / printed page 14

TOPIC 01

Permutations and combinations

Distinguish ordered selections, unordered sets and symmetry in arrangements.

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

01 / Standard#Your turn

r=3

Counts use nonnegative integer sizes; decide whether labels, order and rotations distinguish outcomes.

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.

(r+3)!/(r!2!)

Worked solution

Decide whether choices are alternative cases or successive stages, then check overlap.
Apply the stated relation and retain its conditions.
$N=6!/(3!2!)=60$
Permutations of each identical-letter group do not create new arrangements.

The requested value is 60.

Checks and common pitfalls: Permutations of each identical-letter group do not create new arrangements.

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

Teacher preparation and assessment

Defining relation: Distinguish ordered selections, unordered sets and symmetry in arrangements.
$P_n^r=n!/(n-r)!;\quad C_n^r=n!/[r!(n-r)!]$
Conditions: Counts use nonnegative integer sizes; decide whether labels, order and rotations distinguish outcomes.

Dividing by r! is valid only when every selected set has exactly r! counted orders.