10

10.2 Use Combinations and the Binomial Theorem

VOCABULARY

Combination

A selection of r objects from a group of n objects where the order is not important

Pascal's triangle

An arrangement of the values of _nC_r w a triangular pattern in which each row corresponds to a value of n

Binomial theorem

For any positive integer n, the binomial expansion of (a + b)ⁿ is

(a+ b)ⁿ = _nC₀aⁿb⁰ + _nC₁aⁿ^-1b¹ + _nC₂aⁿ^-2b² + ¼ + _nC_na⁰bⁿ

COMBINATIONS OF n OBJECTS TAKEN r AT A TIME

The number of combinations of r objects taken from a group of n distinct objects is denoted by _nC_r

nCr =

Example 1

Find combinations

Books You are picking 7 books from a stack of 32. If the order of the books you choose is not important, how many different 7 book groups are possible?

32

The number of ways to choose 7 books from 32 is:

7

25

₃₂C₇ =

= 3,365,856

Example 2

Decide to multiply or add combinations

Movie Rentals The local movie rental store is having a special on new releases. The new releases consist of 12 comedies, 8 action, 7 drama, 5 suspense, and 9 family movies.

a. You want exactly 2 comedies and 3 family movies. How many different movie combinations can you rent?

b. You can afford at most 2 movies. How many movie combinations can you rent?

Solution

a.

9

12

You can choose 2 of the 12 comedies and 3 of the 9 family movies. So, the number of possible sets of movies is:

3

6

2

10

₁₂C₂ • ₉C₃ =

= _66 • 84_ = _5544_

b. You can rent 0,1, or 2 movies. Because there are _41_ movies to choose from, the number of possible sets of movies is:

_{_41_}C₀ + _{_41_} C₁ + _{_41_} C₂ = _1 + 41 + 820_ = _862_

Complete the following exercises.

1. Find ₇C₄.

2. Find ₆C₃.

3. Find ₁₂C₁₁.

4. From Example 2, find the number of possible movie combinations if you can choose 2 action movies and 2 dramas.

588

Example 3

Solve a multi-step problem

Reading A popular magazine has 11 articles. You want to read at least 2 of the articles. How many different combinations of articles can you read?

Solution

For each of the 11 articles, you can choose to read or not read the article, so there are _2¹¹_ total combinations. If you read at least _2_ articles, you do not read only a total of _0_ or _1_ articles. So, the number of ways you can read at least 2 articles is:

_2¹¹_ - (₁₁C _{_0_} + ₁₁C _{_1_}) = _2048 - (1 + 11)_ = _2036_.

Complete the following exercise.

5. Your school football team has 10 scheduled games for the season. You want to attend at least 4 games. How many different combinations of games can you attend?

792

PASCAL'S TRIANGLE

The first and last numbers in each row are _1_. Every number other than _1_ is the sum of the closest two numbers in the row directly above it.

Pascal's triangle:	As combinations
n = 0 (0th row)	₀C _{_0_}
n = 1 (1st row)	₁C_{_0_} ₁C_{_1_}
n = 2 (2nd row)	₂C_{_0_} ₂C_{_1_} ₂C_{_2_}
n = 3 (3rd row)	₃C_{_0_} ₃C_{_1_} ₃C_{_2_} ₃C_{_3_}
	As numbers
	_1_
	_1_ _1_
	_1_ _2_ _1_
	_1_ _3_ _3_ _1_

Example 4

Use Pascal's triangle

Class Representatives Out of 5 finalists, your class must choose 3 class representatives. Use Pascal's triangle to find the number of combinations of 3 students that can be chosen as representatives.

Solution

Find ₅C _{_3_} using the 5th row of Pascal's triangle.

n = 5

(5th row)

_1_

_5_

_10_

_5_

_1_

₅C_{_0_}

₅C_{_1_}

₅C_{_2_}

₅C_{_3_}

₅C_{_4_}

₅C_{_1_}

The value of ₅C_{_3_} is the _fourth_ value in the 5th row of Pascal's triangle. Therefore, ₅C_{_3_} = _10_. There are _10_ combinations of class representatives.

Complete the following exercise.

6. In Example 4, use Pascal's triangle to find the number of combinations of 3 students that can be chosen from 8 finalists.

BINOMIAL THEOREM

· For any positive integer n, the binomial expansion of (a + b)ⁿ is:

(a + b)ⁿ = _nC₀aⁿb⁰ + _nC₁aⁿ^-¹b¹ + × × × +_nC_na⁰bⁿ

Notice that each term in the expansion of (a + b)ⁿ has the form _nC_raⁿ^-^1rb^r where r is an integer from 0 to n.

Example 5

Expand a power of a binomial sum

Use the binomial theorem to write the binomial expansion. (x + 4)³

= ₃C₀x³(4)⁰ + ₃C₁x²(4)¹ + ₃C₂x¹(4)² + ₃C₃x⁰(4)³

= _(1)(x³)(1) + (3)(x²)(4) + (3)(x)(16) + (1)(1)(64)_

= _x³ + 12x² + 48x+ 64_

Example 6

Expand a power of a binomial difference

Use the binomial theorem to write the binomial expansion.

(2m - n)⁴ = [2m + (_-n_)]⁴

= ₄C₀(2m)⁴(-n)⁰ + ₄C₁(2m)³(-n)¹ + ₄C₂(2m)²(-n)² + ₄C₃(2m)¹(-n)³ + ₄C₄(2m)⁰(-n)⁴

= _(1)(l6m⁴)(1) + (4)(8m³)(-n) + (6)(4m²)(n²) + (4)(2m)(-n³) + (1)(1)(n⁴)_

= _16m⁴ - 32m³n + 24m²n² - 8mn³ + n⁴_

Example 7

Find a coefficient in an expansion

Find the coefficient of x⁵ in the expansion of (2x - 7)⁹.

Each term in the expansion has the form ₉C_r(2x)⁹^-^r(-7)^r. The term containing x⁵ occurs when r = 4:

₉C₄(2x)⁵(-7)⁴ = _(126)(32x⁵)(2401)_ = _9,680,832x⁵_

The coefficient of x⁵ is _9,680,832_.

Use the binomial theorem to write the binomial expansion.

7. (a + 2b)³

a³ + 6a²b + 12ab + 8b³

8. (6 - s)⁴

1296 - 864s+ 216s² - 24s³ + s⁴

9. Find the coefficient of x⁸ in the expansion of (3x - 2)¹⁰.

1,180,980