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Core Concept

A Permutation is an arrangement of $m$ elements ($m \leq n$) selected from $n$ distinct elements in a specific order.

Key Characteristics

1. Order matters: Different orders count as different permutations
2. No repetition: Each element used at most once
3. Selection: Choose $m$ from $n$ elements ($m \leq n$)

Permutation Formula

General Permutation

The number of permutations of $m$ elements from $n$ distinct elements, denoted $A_n^m$ or $P_n^m$ or $P(n,m)$:

$A_n^m = n(n-1)(n-2)\cdots(n-m+1) = \frac{n!}{(n-m)!}$

Understanding:

• Position 1: $n$ choices
• Position 2: $n-1$ choices
• ...
• Position $m$: $n-m+1$ choices

By multiplication principle: $A_n^m = n \times (n-1) \times \cdots \times (n-m+1)$

Full Permutation

When $m = n$, called full permutation:

$A_n^n = n! = n \times (n-1) \times \cdots \times 2 \times 1$

Convention: $0! = 1$

Common Values

$n$	$n!$
$0$	$1$
$1$	$1$
$2$	$2$
$3$	$6$
$4$	$24$
$5$	$120$
$6$	$720$

Special Permutations

1. Derangement

Number of permutations where no element is in its original position:

$D_n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n\frac{1}{n!}\right)$

Approximation: $D_n \approx \frac{n!}{e}$

2. Circular Permutation

Arranging $n$ distinct elements in a circle:

$Q_n = \frac{A_n^n}{n} = (n-1)!$

(No fixed starting point, divide by $n$)

3. Permutations with Repetition

$n$ elements with $n_1$ identical, $n_2$ identical, ..., $n_k$ identical ($n_1 + n_2 + \cdots + n_k = n$):

$\frac{n!}{n_1! \cdot n_2! \cdots n_k!}$

Calculation Techniques

Technique 1: Step-by-Step Multiplication

Example: Select 3 people from 10 for president, vice-president, and secretary. How many ways?

Solution:

• President: 10 choices
• Vice-president: 9 choices
• Secretary: 8 choices

Answer: $10 \times 9 \times 8 = 720$

Technique 2: Handle Special Elements First

Example: 5 people in a line, person A must be first. How many ways?

Solution:

• A fixed first: 1 way
• Arrange remaining 4: $4! = 24$

Answer: $1 \times 24 = 24$

Technique 3: Complementary Counting

Example: 5 people in a line, A and B not adjacent. How many ways?

Solution:

• Total arrangements: $5! = 120$
• A and B adjacent (treat as one): $2! \times 4! = 48$
• Not adjacent: $120 - 48 = 72$

Answer: $72$

CSCA Practice Problems

[Example 1] Basic (Difficulty ★★☆☆☆)

Calculate $A_6^3$.

Solution:

$A_6^3 = 6 \times 5 \times 4 = 120$

Or

$A_6^3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 120$

Answer: $120$

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[Example 2] Intermediate (Difficulty ★★★☆☆)

5 people line up for a photo, A and B must stand together. How many ways?

Solution:

Grouping method:

1. Treat A and B as one unit, arrange 4 units: $4! = 24$
2. Arrange A and B internally: $2! = 2$

Answer: $24 \times 2 = 48$

Common Misconceptions

❌ Misconception 1: Confusing permutation and combination

Wrong: Not considering order, treating permutation as combination

Correct: Permutation is ordered, combination is unordered

❌ Misconception 2: Forgetting special restrictions

Wrong: Ignoring conditions like "first digit cannot be 0"

Correct: Handle special positions or elements first

Relationship with Combination

$A_n^m = C_n^m \times m!$

Understanding:

• Select $m$ from $n$: $C_n^m$
• Arrange these $m$ elements: $m!$

Study Tips

1. ✅ Understand essence: Permutation emphasizes order
2. ✅ Master formula: $A_n^m = \frac{n!}{(n-m)!}$
3. ✅ Practice techniques: Special elements first, grouping, insertion, complement
4. ✅ Case analysis: Complex problems require classification

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💡 Exam Tip: Permutation is fundamental to combinatorics, mandatory in CSCA! Accounts for about 40% of counting problems. Mastering case analysis and special handling techniques is key.