%%{init: {'flowchart': {'nodeSpacing': 20, 'rankSpacing': 50}}}%%
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
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class T1,T2 task
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PSTAT 5A: Counting
Lecture 7 - Permutations
2025-07-07
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 applications of counting include:
- 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
Simple 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: \(26 \times 26 \times 26 \times 10 \times 10 \times 10 = 26^3 \times 10^3 = 17,576,000\)
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. CBAPermutations of \(n\) Distinct Objects
Key Formula
The number of ways to arrange \(n\) distinct objects is:
\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\]
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
Permutations of \(r\) Objects from \(n\)
Key Formula
\(P(n,r)\) or \(_nP_r\): Number of ways to arrange \(r\) objects selected from \(n\) 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. \(P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336\)
Practice Problem 2
A baseball team has 15 players. How many ways can the coach:
- Arrange all 15 players in a line?
- Choose and arrange 9 players for the starting lineup (batting order matters)?
Solution.
- \(15! = 1,307,674,368,000\)
- \(P(15,9) = \frac{15!}{6!} = 1,816,214,400\)
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. \(\frac{10!}{3! \times 3! \times 1! \times 2! \times 1!} = \frac{3,628,800}{6 \times 6 \times 1 \times 2 \times 1} = \frac{3,628,800}{72} = 50,400\)
