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Section 8-7 Pascal’s Triangle

Notes 8-7

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Pascal's Triangle

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Page 1: Notes 8-7

Section 8-7Pascal’s Triangle

Page 2: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

Page 3: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1

Page 4: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6

Page 5: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6 15

Page 6: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6 15

20

Page 7: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6 15

20 15

Page 8: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6 15

20 15 6

Page 9: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6 15

20 15 6

1

Page 10: Notes 8-7

Warm-upGive the number of subsets of the set of letters {A, E, I, O, U, and Y} that have the given number of

elements.

1. o elements 2. 1 element 3. 2 elements

4. 3 elements 5. 4 elements 6. 5 elements

7. 6 elements

1 6 15

20 15 6

11, 6, 15, 20, 15, 6, 1

Page 11: Notes 8-7

Pascal’s Triangle

Page 12: Notes 8-7

Pascal’s Triangle

The pattern that emerges from displaying rows of combinations

Page 13: Notes 8-7

Pascal’s Triangle

The pattern that emerges from displaying rows of combinations

Let’s build it!

Page 27: Notes 8-7

111

1 2 11 3 3 11 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 1Interactive

Page 29: Notes 8-7

(r + 1)st term in row n in Pascal’s Triangle

To find a particular term in Pascal’s Triangle, let n be the row. Then the value of the (r + 1)st term is nCr.

Interactive

Page 30: Notes 8-7

(r + 1)st term in row n in Pascal’s Triangle

To find a particular term in Pascal’s Triangle, let n be the row. Then the value of the (r + 1)st term is nCr.

Check it out if you don’t believe me.

Interactive

Page 32: Notes 8-7

Properties of Pascal’s Triangle1. The first and last terms of each row are 1’s

Interactive

Page 33: Notes 8-7

Properties of Pascal’s Triangle1. The first and last terms of each row are 1’s

2. The second and next-to-last terms in the nth row are n

Interactive

Page 34: Notes 8-7

Properties of Pascal’s Triangle1. The first and last terms of each row are 1’s

2. The second and next-to-last terms in the nth row are n

3. The rows are symmetrical

Interactive

Page 35: Notes 8-7

Properties of Pascal’s Triangle1. The first and last terms of each row are 1’s

2. The second and next-to-last terms in the nth row are n

3. The rows are symmetrical

4. The sum of the terms in row n is 2n

Interactive

Page 36: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

Page 37: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term:

Page 38: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0

Page 39: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

Page 40: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9!

Page 41: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Page 42: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term:

Page 43: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1

Page 44: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Page 45: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Third term:

Page 46: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Third term: 9C2

Page 47: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Third term: 9C2 = 36

Page 48: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Third term: 9C2 = 36

Fourth term:

Page 49: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Third term: 9C2 = 36

Fourth term: 9C3

Page 50: Notes 8-7

Example 1Use the formula for nCr to find the first four terms in row

9 of Pascal’s Triangle.

First term: 9C0 =

9!9 − 0( )!0!

=9!9! = 1

Second term: 9C1 = 9

Third term: 9C2 = 36

Fourth term: 9C3 = 84

Page 51: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

Page 52: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

We need something to fit the pattern:

Page 53: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

We need something to fit the pattern:1, 3, 6, 10, 15, ...

Page 54: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

We need something to fit the pattern:1, 3, 6, 10, 15, ...

Differences:

Page 55: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

We need something to fit the pattern:1, 3, 6, 10, 15, ...

Differences: 2, 3, 4, 5, ...

Page 56: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

We need something to fit the pattern:1, 3, 6, 10, 15, ...

Differences: 2, 3, 4, 5, ...1, 1, 1, ...

Page 57: Notes 8-7

Example 2Give a polynomial formula for the third term in the nth

row of Pascal’s Triangle.

Interactive

We need something to fit the pattern:1, 3, 6, 10, 15, ...

Differences: 2, 3, 4, 5, ...1, 1, 1, ...

This tells us we are dealing with a 2nd degree polynomial

Page 58: Notes 8-7

Example 2(2, 1), (3, 3), (4, 6), (5, 10)...

Page 59: Notes 8-7

Example 2(2, 1), (3, 3), (4, 6), (5, 10)...

Page 60: Notes 8-7

Example 2(2, 1), (3, 3), (4, 6), (5, 10)...

Page 61: Notes 8-7

Example 2(2, 1), (3, 3), (4, 6), (5, 10)...

Page 62: Notes 8-7

Example 2(2, 1), (3, 3), (4, 6), (5, 10)...

y =12 x2 − 1

2 x

Page 63: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

Page 64: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:

Page 65: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Page 66: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:

Page 67: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1

Page 68: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12

Page 69: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66

Page 70: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220

Page 71: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495

Page 72: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792

Page 73: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924

Page 74: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924 792

Page 75: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924 792 495

Page 76: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924 792 495 220

Page 77: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924 792 495 220 66

Page 78: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924 792 495 220 66 12

Page 79: Notes 8-7

Example 3Here are the first 6 terms of row 11 of Pascal’s Triangle:

1, 11, 55, 165, 330, 462.Use them to construct row 12.

There are 12 terms in row 11, so the entire row is:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Row 12:1 12 66 220 495 792 924 792 495 220 66 12 1

Page 80: Notes 8-7

Homework

Page 81: Notes 8-7

Homework

p. 532 #1-20