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11.1 – Pascal’s Triangle and the Binomial Theorem

11.1 – Pascal’s Triangle and the Binomial Theorem

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Page 1: 11.1 – Pascal’s Triangle and the Binomial Theorem

11.1 – Pascal’s Triangle and the Binomial Theorem

Page 2: 11.1 – Pascal’s Triangle and the Binomial Theorem

FACTORIALS, COMBINATIONS

Factorial is denoted by the symbol “!”. The factorial of a number is calculated by multiplying all integers from the number to 1.

Formal DefinitionThe symbol n!, is define as the product of all the integers from n to 1. In other words,

n! = n(n - 1)(n – 2)(n – 3) · · · 3 · 2 · 1

Also note that by definition,0! = 1

Example #9

3! 3 2 1 6(9 3)! 6! 6 5 4 3 2 1 7209! 9 8 7 6 5 4 3 2 1 362,880

Page 3: 11.1 – Pascal’s Triangle and the Binomial Theorem

Combinations

Definition

Combinations give the number of ways x element can be selected from n distinct elements. The total number of combinations is given by,

and is read as “the number of combinations of n elements selected x at a time.”

The formula for the number of combinations for selecting x from n distinct elements is,

Note:

n xC

!!( )!n x

nC

x n x

! ! !

!( )! ! 0!n n

n n nC

n n n n

!n1 0

! ! !0!( 0)! 0! !n

n n nC

n n

!n1

Page 4: 11.1 – Pascal’s Triangle and the Binomial Theorem

Combinations

Example #10

2

5 3

7 4

4 0

3 3

5! 5! 5 3 2 14 103!(5 3)! 3! 2! 3 2 1 2 1

7! 7! 7 6 5 4 3 2 135

4!(7 4)! 4! 3! 4 3 2 1 3 2 1

1

1

C

C

C

C

Page 5: 11.1 – Pascal’s Triangle and the Binomial Theorem

The Binomial Theorem

Strategy only: how do we expand these?

1. (x + 2)2 2. (2x + 3)2

3. (x – 3)3 4. (a + b)4

Page 6: 11.1 – Pascal’s Triangle and the Binomial Theorem

The Binomial TheoremSolutions

1. (x + 2)2 = x2 + 2(2)x + 22 = x2 + 4x + 4

2. (2x + 3)2 = (2x)2 + 2(3)(2x) + 32 = 4x2 + 12x + 9

3. (x – 3)3 = (x – 3)(x – 3)2 = (x – 3)(x2 – 2(3)x + 32) =(x – 3)(x2 – 6x + 9) = x(x2 – 6x + 9) – 3(x2 – 6x + 9) =x3 – 6x2 + 9x – 3x2 + 18x – 27 = x3 – 9x2 + 27x – 27

4. (a + b)4 = (a + b)2(a + b)2 = (a2 + 2ab + b2)(a2 + 2ab + b2) =a2(a2 + 2ab + b2) + 2ab(a2 + 2ab + b2) + b2(a2 + 2ab + b2) =a4 + 2a3b + a2b2 + 2a3b + 4a2b2 + 2ab3 + a2b2 + 2ab3 + b4 =a4 + 4a3b + 6a2b2 + 4ab3 + b4

Page 7: 11.1 – Pascal’s Triangle and the Binomial Theorem

THAT is a LOT of work!

Isn’t there an easier way?

Page 8: 11.1 – Pascal’s Triangle and the Binomial Theorem

Introducing: Pascal’s Triangle

Take a moment to copy the first 6 rows. What patterns do you see?

Row 5Row 6

Page 9: 11.1 – Pascal’s Triangle and the Binomial Theorem

Use Pascal’s Triangle to expand (a + b)5.

The Binomial Theorem

Use the row that has 5 as its second number.

In its simplest form, the expansion is a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.

The exponents for a begin with 5 and decrease.

     1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5

The exponents for b begin with 0 and increase.

Row 5

Page 10: 11.1 – Pascal’s Triangle and the Binomial Theorem

Use Pascal’s Triangle to expand (x – 3)4.

The Binomial Theorem

First write the pattern for raising a binomial to the fourth power.

1 4 6 4 1   Coefficients from Pascal’s Triangle.

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Since (x – 3)4 = (x + (–3))4, substitute x for a and –3 for b.

(x + (–3))4 = x4 + 4x3(–3) + 6x2(–3)2 + 4x(–3)3 + (–3)4

= x4 – 12x3 + 54x2 – 108x + 81

The expansion of (x – 3)4 is x4 – 12x3 + 54x2 – 108x + 81.

Page 11: 11.1 – Pascal’s Triangle and the Binomial Theorem

The Binomial Theorem

nrrnnnnnn bbarnr

nba

n

nba

n

nba

n

naba

...

)!(!

!...

)!3(!3

!

)!2(!2

!

)!1(!1

!)( 33221

For any positive integer, n

!!( )!n x

nC

x n x

Page 12: 11.1 – Pascal’s Triangle and the Binomial Theorem

The Binomial TheoremUse the Binomial Theorem to expand (x – y)9.

Write the pattern for raising a binomial to the ninth power.(a + b)9 = 9C0a9 + 9C1a8b + 9C2a7b2 + 9C3a6b3

+ 9C4a5b4 + 9C5a4b5 + 9C6a3b6 + 9C7a2b7 + 9C8ab8 + 9C9b9

Substitute x for a and –y for b. Evaluate each combination.(x – y)9 = 9C0x9 + 9C1x8(–y) + 9C2x7(–y)2 + 9C3x6(–y)3

+ 9C4x5(–y)4 + 9C5x4(–y)5 + 9C6x3(–y)6 + 9C7x2(–y)7 + 9C8x(–y)8 + 9C9(–y)9

= x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9

The expansion of (x – y)9 is x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4

– 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9.

Page 13: 11.1 – Pascal’s Triangle and the Binomial Theorem

Let’s Try SomeExpand the following

a) (x-y5)3 b) (3x-2y)4

Page 14: 11.1 – Pascal’s Triangle and the Binomial Theorem

Let’s Try SomeExpand the following

(x-y5)3

Page 15: 11.1 – Pascal’s Triangle and the Binomial Theorem

Let’s Try SomeExpand the following

(3x-2y)4

Page 16: 11.1 – Pascal’s Triangle and the Binomial Theorem

Let’s Try SomeExpand the following(3x-2y)4

Page 17: 11.1 – Pascal’s Triangle and the Binomial Theorem

How does this relate to probability?

• You can use the Binomial Theorem to solve probability problems. If an event has a probability of success p and a probability of failure q, each term in the expansion of (p + q)n represents a probability.

Example: 10C2 * p8 q2 represents the probability of 8 successes in 10 tries

Page 18: 11.1 – Pascal’s Triangle and the Binomial Theorem

The Binomial TheoremBrianna makes about 90% of the shots on goal she attempts. Find the probability that Bri makes exactly 7 out of 12 consecutive goals.

Since you want 7 successes (and 5 failures), use the term p7q5. This term has the coefficient 12C5.

Probability (7 out of 10) = 12C5 p7q5

= 0.0037881114 Simplify.

Bri has about a 0.4% chance of making exactly 7 out of 12 consecutive goals.

= • (0.9)7(0.1)5 The probability p of success = 90%, or 0.9. 12! 5! •7!