04/20/23 MATH 106, Section 8 1

Section 8

Pascal’s Triangle

Questions about homework?Submit homework!

04/20/23 MATH 106, Section 8 2

How many ways can we choose a committee of 5 from a class of one instructor and 13 students if

the committee may consist of any 5 of the 14 individuals.

the committee must include the instructor.

the committee must not include the instructor.

C(14,5) = 2002

C(13,4) = 715

C(13,5) = 1287

Why do these two answers sum to the previous answer?

#1

04/20/23 MATH 106, Section 8 3

Pascal’s Formula• Pascal observed the following:

C(n, k) = C(n-1, k-1) + C(n-1, k)

The number of ways to choose k items from the n items

The number of ways to choose k–1 items from the n–1 items

Set aside 1 “special” item leaving n–1 items. Select k items from the n items as follows:

The number of ways to choose k items from the n–1 items

Choose the “special” item and then choose k–1 items from the n–1 items

or

Do not choose the “special” item and then choose k items from the n–1 items

04/20/23 MATH 106, Section 8 4

Pascal’s Formula• Pascal observed the following:

C(n, k) = C(n-1, k-1) + C(n-1, k)

• Let’s check it!

C(5, 3) =

C(7, 2) =

C(6, 1) =

C( ) + C( )4, 2 4, 3

10 6 4

C( ) + C( )6, 1 6, 2

21 6 15

C( ) + C( )5, 0 5, 1

6 1 5

04/20/23 MATH 106, Section 8 5

Pascal’s Triangle

C(0, 0)

C(1, 1)C(1, 0)

C(2, 0) C(2, 2)C(2, 1)

C(3, 0) C(3, 2)C(3, 1) C(3, 3)

C(4, 0) C(4, 2)C(4, 1) C(4, 3) C(4, 4)

C(5, 0) C(5, 2)C(5, 1) C(5, 3) C(5, 4) C(5, 5)

How do we compute the values?How do we compute the values?

Note: The first row is row 0 (that is, n is 0).Note: The first row is row 0 (that is, n is 0).

04/20/23 MATH 106, Section 8 6

Pascal’s Triangle

C(0, 0)

C(1, 1)C(1, 0)

C(2, 0) C(2, 2)C(2, 1)

C(3, 0) C(3, 2)C(3, 1) C(3, 3)

C(4, 0) C(4, 2)C(4, 1) C(4, 3) C(4, 4)

C(5, 0) C(5, 2)C(5, 1) C(5, 3) C(5, 4) C(5, 5)1

1 1

1 12

1 3 3 1

1 4 6 4 1

04/20/23 MATH 106, Section 8 7

Pascal’s Triangle1

1 1

1 12

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

#2

In the Pascal’s Triangle displayed below, add the entries up to and including those corresponding to the row for n =10.

04/20/23 MATH 106, Section 8 8

We now have two ways to calculate C(n, k)

• The formula from Section 5

• Pascal’s Triangle

Use Pascal’s Triangle to evaluate each of the following:

C(6, 3) C(10, 4)

C(7, 5) C(10, 6)

C(8, 4) C(10, 5)

20 210

21 210

70 252

#3

04/20/23 MATH 106, Section 8 9

Homework Hints:In Section 8 Homework Problem #5,

In Section 8 Homework Problem #9(b),

In Section 8 Homework Problem #10(b),

you can choose to construct the triangle from the beginning, or you can choose to copy the partially constructed triangle in Section 8 of the textbook (after verifying that it is correct) and then add the required rows to complete the triangle.

you can use your calculation in part (a) together with the GOOD = ALL BAD principle.

notice that there is only one way to alternate speeches and musical selection, after which the order of the speeches and the order of the musical selections must be decided.

04/20/23 MATH 106, Section 8 10

Section 9

Binomial Expansion

04/20/23 MATH 106, Section 8 11

Let’s take a look back …

• Section 7, homework problem #6 0 7 C(7, 0)

1 6 C(7, 1)

2 5 C(7, 2)

3 4 C(7, 3)

4 3 C(7, 4)

5 2 C(7, 5)

6 1 C(7, 6)

7 0 C(7, 7)

H T

• A coin is flipped 7 times. What percent of the times will result in each of the possibilities?

• You must calculate the combinations for each number of heads

• One each flip, there are two possible outcomes.

• How did we figure the percentages?

These match the row of Pascal’s Triangle corresponding to n = 7

04/20/23 MATH 106, Section 8 12

What is “Binomial Expansion”?

Take an expression of the form (x + y)n and multiply it out for n = 2 and for n = 3.

(x + y)2 =

Before simplification, how many terms did we get?

(x + y)3 =

Before simplification, how many terms did we get?

(x + y)(x + y) = x2 + yx +xy + y2 = x2 + 2xy + y2

4 = 22

(x + y)(x + y)(x + y) =x3 + xxy + xyx +xyy +yxx + yxy +yyx +y3 = x3 + 3x2y + 3xy2 + y3

8 = 23

RECIPE FOR CHOOSING ONE TERM IN THE EXPANSION OF (x + y)n BEFORE SIMPLIFICATION:

Choose x or y from first factor and then choose x or y from second factor and then …

#1

04/20/23 MATH 106, Section 8 13

Without doing any multiplication, find the following expansion:

(x + y)7 =

If we did the multiplication, how many terms would we get before simplification?

x7 + x6y + x5y2 + x4y3 + x3y4 + x2y5 + xy6 + y77 21 35 35 21 7

128 = 27

RECIPE FOR CHOOSING ONE TERM IN THE EXPANSION OF (x + y)7 BEFORE SIMPLIFICATION:

Choose x or y from first factor and then choose x or y from second factor and then …

Observe that the coefficients in the simplified expansion of (x + y)7 match the row of Pascal’s Triangle corresponding to 7.

#2

04/20/23 MATH 106, Section 8 14

It’s the same idea as flipping a coin 7 times…

• Each time we compute a term in the expansion of (x + y)7, we choose x or y seven times in a row.

• Each time flip a coin seven times in a row, 7 choices of H or T are made in a row

• We can use Pascal’s triangle …

0 7 C(7, 0)

1 6 C(7, 1)

2 5 C(7, 2)

3 4 C(7, 3)

4 3 C(7, 4)

5 2 C(7, 5)

6 1 C(7, 6)

7 0 C(7, 7)

y x

(x + y)7 = x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

Notice the symmetry!

(x + y)7 = x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

Notice the symmetry!

04/20/23 MATH 106, Section 8 15

The Binomial Theorem

(x + y)n = C(n, 0)xny0 + C(n, 1)xn-1y1 + C(n, 2)xn-2y2 + C(n, 3)xn-3y3 +… + C(n, n-2)x2yn-2 + C(n, n-1)x1yn-1 + C(n, n)x0yn

(x + y)n = C(n, 0)xny0 + C(n, 1)xn-1y1 + C(n, 2)xn-2y2 + C(n, 3)xn-3y3 +… + C(n, n-2)x2yn-2 + C(n, n-1)x1yn-1 + C(n, n)x0yn

(bottom of page 67 in the textbook)

(x + y)n = C(n, n)xny0 + C(n, n-1)xn-1y1 + C(n, n-2)xn-2y2 + C(n, n-3)xn-3y3

+ … + C(n, 2)x2yn-2 + C(n, 1)x1yn-1 + C(n, 0)x0yn

(x + y)n = C(n, n)xny0 + C(n, n-1)xn-1y1 + C(n, n-2)xn-2y2 + C(n, n-3)xn-3y3

+ … + C(n, 2)x2yn-2 + C(n, 1)x1yn-1 + C(n, 0)x0yn

Because of symmetry, we could choose to write the formula in the Binomial Theorem as follows: