Learn on PengiSaxon Algebra 2Chapter 5: Lessons 41-50, Investigation 5

Lesson 42: Finding Permutations and Combinations (Exploration: Pascal's Triangle and Combinations)

In this Grade 10 Saxon Algebra 2 lesson, students learn to calculate factorials, permutations using the formula P(n, r) = n!/(n-r)!, and combinations using C(n, r) = n!/[r!(n-r)!], including distinguishable permutations with repeated elements. The lesson also covers Pascal's Triangle and its connection to combinations. Students practice determining when order matters to decide whether a permutation or combination applies to a given problem.

Section 1

πŸ“˜ Finding Permutations and Combinations

New Concept

A permutation is a selection of items where order is important.

Why it matters

Mastering permutations and combinations is your first step into combinatorics, the art of sophisticated counting. This skill is crucial for probability, computer science algorithms, and making strategic decisions in complex scenarios.

What’s next

Next, you’ll learn the formal notation and formulas to precisely calculate the number of possible arrangements and selections.

Section 2

Factorial

The factorial of a positive integer nn, denoted as n!n!, is the product of all positive integers up to and including nn. By definition, zero factorial is 1.

n!=n(nβˆ’1)(nβˆ’2)…1n! = n(n - 1)(n - 2) \ldots 1
0!=10! = 1

Evaluate 4!4!: 4!=4Γ—3Γ—2Γ—1=244! = 4 \times 3 \times 2 \times 1 = 24.
Evaluate 8!6!\frac{8!}{6!}: 8Γ—7Γ—6!6!=8Γ—7=56\frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56.
Evaluate 7!3!(7βˆ’3)!\frac{7!}{3!(7-3)!}: 7!3!4!=7Γ—6Γ—5Γ—4!3Γ—2Γ—1Γ—4!=2106=35\frac{7!}{3!4!} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = \frac{210}{6} = 35.

Think of a factorial as a countdown multiplication party! To find 5!5!, you just multiply 5 by every whole number below it until you hit 1. It’s a super quick way to find the total possible arrangements of a full set of distinct items. Don't forget the weird guest at the party: 0!0! is always 1!

Section 3

Permutations

A permutation is an arrangement of objects where the order is important. The number of permutations of nn objects taken rr at a time is:

P(n,r)=n!(nβˆ’r)!P(n, r) = \frac{n!}{(n - r)!}
For permutations with repeated items, the formula is n!q1!β‹…q2!β‹―qk!\frac{n!}{q_1! \cdot q_2! \cdots q_k!}.

How many ways can a President and VP be chosen from 8 candidates? P(8,2)=8!(8βˆ’2)!=8!6!=56P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 56.
Find the distinguishable permutations of the letters in 'CHEESE': 6!3!=7206=120\frac{6!}{3!} = \frac{720}{6} = 120.
How many 3-digit lock codes can be made from digits 1-7 if no digit repeats? P(7,3)=7!(7βˆ’3)!=210P(7, 3) = \frac{7!}{(7-3)!} = 210.

Permutations are all about position! Think of it like a race: getting 1st, 2nd, and 3rd place is a different outcome than the same three people in a different order. Use this when arranging things where the sequence is critical, like letters in a password or runners finishing a race. Order is everything here!

Section 4

Combinations

A combination is a selection of items where order does not matter. The number of combinations of nn objects taken rr at a time is:

C(n,r)=n!r!(nβˆ’r)!C(n, r) = \frac{n!}{r!(n - r)!}

How many 3-topping pizzas can be made from 8 available toppings? C(8,3)=8!3!(8βˆ’3)!=56C(8, 3) = \frac{8!}{3!(8-3)!} = 56.
How many ways can you choose a 4-person team from 10 students? C(10,4)=10!4!(10βˆ’4)!=210C(10, 4) = \frac{10!}{4!(10-4)!} = 210.
How many 5-card hands can be drawn from a 52-card deck? C(52,5)=52!5!47!=2,598,960C(52, 5) = \frac{52!}{5!47!} = 2,598,960.

Combinations are for when you just want a group, and the order you pick them in is irrelevant. Think of making a fruit salad: putting in an apple then a banana is the same salad as a banana then an apple. Use this when you're selecting a committee, a hand of cards, or pizza toppings!

Book overview

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Chapter 5: Lessons 41-50, Investigation 5

  1. Lesson 1

    Lesson 41: Using the Pythagorean Theorem and the Distance Formula (Exploration: Visualizing the Pythagorean Theorem)

  2. Lesson 2

    LAB 7: Graphing Calculator: Calculating Permutations and Combinations

  3. Lesson 3Current

    Lesson 42: Finding Permutations and Combinations (Exploration: Pascal's Triangle and Combinations)

  4. Lesson 4

    Lesson 43: Solving Systems of Linear Inequalities

  5. Lesson 5

    Lesson 44: Rationalizing Denominators

  6. Lesson 6

    LAB 8: Graphing Calculator: Applying Linear and Median Regression

  7. Lesson 7

    Lesson 45: Finding the Line of Best Fit (Exploration: Collecting and Analyzing Data)

  8. Lesson 8

    Lesson 46: Finding Trigonometric Functions and their Reciprocals

  9. Lesson 9

    Lesson 47: Graphing Exponential Functions

  10. Lesson 10

    Lesson 48: Understanding Complex Fractions

  11. Lesson 11

    Lesson 49: Using the Binomial Theorem

  12. Lesson 12

    Lesson 50: Finding Inverses of Relations and Functions (Exploration: Graphing a Function and its Inverse)

  13. Lesson 13

    Investigation 5: Finding the Binomial Distribution

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

πŸ“˜ Finding Permutations and Combinations

New Concept

A permutation is a selection of items where order is important.

Why it matters

Mastering permutations and combinations is your first step into combinatorics, the art of sophisticated counting. This skill is crucial for probability, computer science algorithms, and making strategic decisions in complex scenarios.

What’s next

Next, you’ll learn the formal notation and formulas to precisely calculate the number of possible arrangements and selections.

Section 2

Factorial

The factorial of a positive integer nn, denoted as n!n!, is the product of all positive integers up to and including nn. By definition, zero factorial is 1.

n!=n(nβˆ’1)(nβˆ’2)…1n! = n(n - 1)(n - 2) \ldots 1
0!=10! = 1

Evaluate 4!4!: 4!=4Γ—3Γ—2Γ—1=244! = 4 \times 3 \times 2 \times 1 = 24.
Evaluate 8!6!\frac{8!}{6!}: 8Γ—7Γ—6!6!=8Γ—7=56\frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56.
Evaluate 7!3!(7βˆ’3)!\frac{7!}{3!(7-3)!}: 7!3!4!=7Γ—6Γ—5Γ—4!3Γ—2Γ—1Γ—4!=2106=35\frac{7!}{3!4!} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = \frac{210}{6} = 35.

Think of a factorial as a countdown multiplication party! To find 5!5!, you just multiply 5 by every whole number below it until you hit 1. It’s a super quick way to find the total possible arrangements of a full set of distinct items. Don't forget the weird guest at the party: 0!0! is always 1!

Section 3

Permutations

A permutation is an arrangement of objects where the order is important. The number of permutations of nn objects taken rr at a time is:

P(n,r)=n!(nβˆ’r)!P(n, r) = \frac{n!}{(n - r)!}
For permutations with repeated items, the formula is n!q1!β‹…q2!β‹―qk!\frac{n!}{q_1! \cdot q_2! \cdots q_k!}.

How many ways can a President and VP be chosen from 8 candidates? P(8,2)=8!(8βˆ’2)!=8!6!=56P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 56.
Find the distinguishable permutations of the letters in 'CHEESE': 6!3!=7206=120\frac{6!}{3!} = \frac{720}{6} = 120.
How many 3-digit lock codes can be made from digits 1-7 if no digit repeats? P(7,3)=7!(7βˆ’3)!=210P(7, 3) = \frac{7!}{(7-3)!} = 210.

Permutations are all about position! Think of it like a race: getting 1st, 2nd, and 3rd place is a different outcome than the same three people in a different order. Use this when arranging things where the sequence is critical, like letters in a password or runners finishing a race. Order is everything here!

Section 4

Combinations

A combination is a selection of items where order does not matter. The number of combinations of nn objects taken rr at a time is:

C(n,r)=n!r!(nβˆ’r)!C(n, r) = \frac{n!}{r!(n - r)!}

How many 3-topping pizzas can be made from 8 available toppings? C(8,3)=8!3!(8βˆ’3)!=56C(8, 3) = \frac{8!}{3!(8-3)!} = 56.
How many ways can you choose a 4-person team from 10 students? C(10,4)=10!4!(10βˆ’4)!=210C(10, 4) = \frac{10!}{4!(10-4)!} = 210.
How many 5-card hands can be drawn from a 52-card deck? C(52,5)=52!5!47!=2,598,960C(52, 5) = \frac{52!}{5!47!} = 2,598,960.

Combinations are for when you just want a group, and the order you pick them in is irrelevant. Think of making a fruit salad: putting in an apple then a banana is the same salad as a banana then an apple. Use this when you're selecting a committee, a hand of cards, or pizza toppings!

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 5: Lessons 41-50, Investigation 5

  1. Lesson 1

    Lesson 41: Using the Pythagorean Theorem and the Distance Formula (Exploration: Visualizing the Pythagorean Theorem)

  2. Lesson 2

    LAB 7: Graphing Calculator: Calculating Permutations and Combinations

  3. Lesson 3Current

    Lesson 42: Finding Permutations and Combinations (Exploration: Pascal's Triangle and Combinations)

  4. Lesson 4

    Lesson 43: Solving Systems of Linear Inequalities

  5. Lesson 5

    Lesson 44: Rationalizing Denominators

  6. Lesson 6

    LAB 8: Graphing Calculator: Applying Linear and Median Regression

  7. Lesson 7

    Lesson 45: Finding the Line of Best Fit (Exploration: Collecting and Analyzing Data)

  8. Lesson 8

    Lesson 46: Finding Trigonometric Functions and their Reciprocals

  9. Lesson 9

    Lesson 47: Graphing Exponential Functions

  10. Lesson 10

    Lesson 48: Understanding Complex Fractions

  11. Lesson 11

    Lesson 49: Using the Binomial Theorem

  12. Lesson 12

    Lesson 50: Finding Inverses of Relations and Functions (Exploration: Graphing a Function and its Inverse)

  13. Lesson 13

    Investigation 5: Finding the Binomial Distribution