PSTAT 5A: Counting

Lecture 7 - Permutations

Narjes Mathlouthi

2025-07-07

Welcome to Lecture 7

The art and science of systematic enumeration

Today’s Learning Objectives

Learning Objectives

By the end of this lecture, you will be able to:

Why Study Counting?

Counting helps us:

  • Calculate probabilities for complex events
  • Solve optimization problems
  • Understand combinations in genetics, computer science
  • Analyze algorithms and data structures
  • Make decisions involving arrangements and selections

Real-world Examples:

  • Cryptography: Password strength and encryption key space
  • Genetics: DNA sequence analysis and gene combinations
  • Tournament brackets: March Madness and sports competitions
  • Lottery odds: Probability calculations for games of chance
  • Password security: Character combinations and brute force protection

The Counting Principle

Multiplication Rule

If a procedure consists of k separate tasks where:

  • Task 1 can be performed in n_1 ways

  • Task 2 can be performed in n_2 ways

  • …

  • Task k can be performed in n_k ways

Then, the entire procedure can be performed in n_1 \times n_2 \times \cdots \times n_k ways

Visualization

%%{init: {'flowchart': {'nodeSpacing': 5, 'rankSpacing': 20}}}%%
flowchart TD
    Start([🟢 Start]) --> T1[📋 Task 1<br/>n₁ ways]
    T1 --> C1[Choice 1]
    T1 --> C2[Choice 2]
    T1 --> Cn1[Choice n₁]
    
    C1 --> T2[📋 Task 2<br/>n₂ ways]
    C2 --> T2
    Cn1 --> T2
    
    T2 --> C21[Choice 1]
    T2 --> C22[Choice 2]
    T2 --> C2n[Choice n₂]
    
    C21 --> Total[🎯 Total ways<br/>n₁ × n₂ × ... × nₖ]
    C22 --> Total
    C2n --> Total
    
    classDef start fill:#d4edda,stroke:#155724,stroke-width:3px
    classDef task fill:#d1ecf1,stroke:#0c5460,stroke-width:2px
    classDef choice fill:#fff3cd,stroke:#856404,stroke-width:1px
    classDef total fill:#f8d7da,stroke:#721c24,stroke-width:3px
    
    class Start start
    class T1,T2 task
    class C1,C2,Cn1,C21,C22,C2n choice
    class Total total

Counting Example

Format: ABC-123

\underbrace{A \; B \; \_ \; \ - \_ \; \_ \; \_}_{positions}

  • First position: 26 letters

  • Second position: 26 letters

  • Third position: 26 letters

  • Fourth position: 10 digits

  • Fifth position: 10 digits

  • Sixth position: 10 digits

Solution. Total possibilities: 26 \times 26 \times 26 \times 10 \times 10 \times 10 = 26^3 \times 10^3

Total possibilities = 17,576,000

Restaurant Menu Example

A restaurant offers:

🍤 Appetizers: 4

🍲 Main Courses: 6

🍰 Desserts: 3

How many different three-course meals are possible?

Solution. 4 \times 6 \times 3 = 72 different meals

Practice Problem 1

A password must contain:

  • Exactly 8 characters

  • Each character is either a letter (26 possibilities) or digit (10 possibilities)

How many possible passwords are there?

Solution. Each position has 26 + 10 = 36 choices.

Total: 36^8 = 2,821,109,907,456 passwords

What Are Permutations?

Permutation

An arrangement of objects where order matters

Order Matters

  • Race finish positions (1st, 2nd, 3rd)

  • Seating arrangements

  • Passwords

  • DNA sequences

All permutations of ABC:
1. ABC
2. ACB
3. BAC
4. BCA
5. CAB
6. CBA

Permutations of n Distinct Objects

Key Formula

The number of ways to arrange n distinct objects is:

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

Note

Seating Process:

1st position: 5 choices (Alice, Bob, Carol, David, Eve)

2nd position: 4 choices (whoever is left)

3rd position: 3 choices (whoever is left)

4th position: 2 choices (whoever is left)

5th position: 1 choice (last person)

How many ways can 5 people sit in a row?

Solution. 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 ways

Factorial Values

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

Note

0! = 1 by definition

Arranging with Order

Key Formula

P(n,r) or _nP_r: Number of ways to arrange r objects (a subset) selected from n (total num.) distinct objects

P(n,r) = \frac{n!}{(n-r)!}

Note

P_{k,n} = \frac{n!}{(n-k)!} =

= \frac{ n(n-1)\cdots(n-k+1)\, \overbrace{(n-k)(n-k-1)\cdots3\cdot2\cdot1}^{(n-k)!} }{ (n-k)! }

= \underbrace{ n (n-1) (n-2) \cdots (n-k+1) }_{k \text{ terms}}

Fill in k slots with no repetitions
\underbrace{n \; (n-1) \; \_ \; \_ \; \cdots}_{k}

Note that if we allowed repetitions we would get n^k \underbrace{n \; n \; n \; \cdots \; n}_{k}

Example

How many ways can we select and arrange 3 people from a group of 8 for president, vice-president, and secretary?

Solution. P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336

Understanding P(n,r)

Why is P(n,r) = \frac{n!}{(n-r)!}?

  • First position: n choices
  • Second position: (n-1) choices
  • Third position: (n-2) choices
  • …
  • r-th position: (n-r+1) choices

Total: n \times (n-1) \times (n-2) \times \cdots \times (n-r+1) = \frac{n!}{(n-r)!}

Practice Problem 2

A baseball team has 15 players. How many ways can the coach:

  1. Arrange all 15 players in a line?
  2. Choose and arrange 9 players for the starting lineup (batting order matters)?

Solution.

  1. 15! = 1,307,674,368,000
  2. P(15,9) = \frac{15!}{6!} = 1,816,214,400

Permutations with Repetition

Permutations with Repetition

When some objects are identical, we have fewer distinct arrangements

If we have n objects where:

  • n_1 are of type 1

  • n_2 are of type 2

  • …

  • n_k are of type k

Number of distinct arrangements: \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}

Permutations with Repetition Example

How many distinct arrangements are there of the letters in “STATISTICS”?

Tip

S-T-A-T-I-S-T-I-C-S

  • Total letters: 10

  • S appears 3 times

  • T appears 3 times

  • A appears 1 time

  • I appears 2 times

  • C appears 1 time

Solution. \frac{10!}{3! \times 3! \times 1! \times 2! \times 1!} = \frac{3,628,800}{6 \times 6 \times 1 \times 2 \times 1} = \frac{3,628,800}{72} = 50,400

Learning Objectives Summary

What We’ve Covered

In this lecture, we’ve addressed all the learning objectives:

Questions?

Office Hours: Thursday’s 11 AM On Zoom (Link on Canvas)

Email: nmathlouthi@ucsb.edu

Next Class: Counting continued

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