7 further topics in algebra © 2008 pearson addison-wesley. all rights reserved sections 7.4–7.7

7

Further Topics in Algebra

© 2008 Pearson Addison-Wesley.All rights reserved

Sections 7.4–7.7

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-2

7.4 The Binomial Theorem

7.5 Mathematical Induction

7.6Counting Theory

7.7Basics of Probability

Further Topics in Algebra7


The Binomial Theorem7.4A Binomial Expansion Pattern ▪ Pascal’s Triangle ▪ n-factorial ▪ Binomial Coefficients ▪ The Binomial Theorem ▪ kth Term of a Binomial Expansion


Evaluate each binomial coefficient.

7.4 Example 1 Evaluating Binomial Coefficients (page 686)


7.4 Example 1 Evaluating Binomial Coefficients (cont.)

Graphing calculators calculate binomial coefficients using the notation nCr in the MATH menu.


Write the binomial expansion of (x + y)8.

7.4 Example 2 Applying the Binomial Theorem (page 688)


7.4 Example 2 Applying the Binomial Theorem (cont.)


7.4 Example 5 Finding a Particular Term of a Binomial Expansion (page 690)

Find the fourth term of (2c – d)12.

Use the formula with n = 12, k = 4,

k – 1 = 3, and n – (k – 1) = 9.

The fourth term of the expansion is


Mathematical Induction7.5Proof by Mathematical Induction ▪ Proving Statements ▪ Generalized Principle of Mathematical Induction ▪ Proof of the Binomial Theorem


7.5 Example 1 Proving an Equality Statement (page 693)

Let Sn be the statement

Prove that Sn is true for every possible integer n.

Step 1: Verify the statement for n = 1.

which is true.

If n = 1, S1 becomes


7.5 Example 1 Proving an Equality Statement (cont.)

Step 2: Show that Sk implies Sk+1 ,where

Assume that Sk is a true statement.

Add the (k + 1)st term to both sides of the equation to obtain Sk+1.



The final result is the statement for n = k + 1. If Sk is true, then Sk+1 is also true.

The two steps required for a proof by mathematical induction have been completed, so the statement Sn is true for every positive integer value of n.


7.5 Example 2 Proving an Inequality Statement (page 694)

Prove Assume a and b are constants,

with b ≠ 0.

Step 1: Verify the statement for n = 1.

which is true.

If n = 1 and b ≠ 0, S1 is


7.5 Example 2 Proving an Inequality Statement (cont.)

Step 2: Show that Sk implies Sk+1 ,where Sk is the statement

Sk+1 is the statement

Assume that Sk is a true statement. Multiply both sides of the equation by the (k + 1)st term to obtain Sk+1.


7.5 Example 2 Proving an Inequality Statement (cont.)




7.5 Example 3 Using the Generalized Principle (page 694)

Let Sn represent the statement 4n > 4n. Show that Sn is true for all values of n such that n ≥ 2.

Step 1: Verify that the statement is false for n = 1 and true for n = 2.

If n = 1, S1 is which is false.

If n = 2, S2 is which is true.


7.5 Example 3 Using the Generalized Principle (cont.)

Step 2: Show that Sk implies Sk+1 ,where

Assume that Sk is a true statement. Multiply both sides of the equation by 4.

Rewrite 16k as 4k + 12k.


7.5 Example 3 Using the Generalized Principle (cont.)

Since k > 2, 12k > 24, which implies that 12k > 4. Thus,




Counting Theory7.6Fundamental Principle of Counting ▪ Permutations ▪ Combinations ▪ Distinguishing Between Permutations and Combinations


7.6 Example 1 Using the Fundamental Principle of Counting

(page 699)

Alexia is shopping for a desktop computer system. She has found 6 computers, 4 monitors, and 3 printers that meet her needs. Find the number of possible computer systems that can be assembled from these choices.

Using the fundamental principle of counting, there are 6 ∙ 4 ∙ 3 = 72 possible systems.


7.6 Example 2 Using the Fundamental Principle of Counting

(page 700)

Ethel has 9 different pictures that she wants to hang in a row on a wall in her new retirement apartment. How many different arrangements of the pictures are possible?

Using the fundamental principle of counting, there are 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 362,880 different arrangements.


7.6 Example 3 Arranging r of n Items (r < n) (page 700)

Ethel wants to hang 6 out of 9 pictures in a row on a wall in her new retirement apartment. How many different arrangements of the pictures are possible?

Using the fundamental principle of counting, there are 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 = 60,480 different arrangements.


7.6 Example 4 Using the Permutations Formula (page 701)

Find each value.

(a) The number of permutations of the digits 0, 2, 4, 6 and 8.

The digits can be arranged in 120 ways.



(b) The number of permutations of 4 of the digits 0, 2, 4, 6 and 8.

Four of the five digits can be arranged in 120 ways.



A club with 25 members is electing 4 officers: a president, a vice-president, a secretary, and a treasurer. In how many different ways can the 4 officers be chosen?

Using the fundamental principle of counting, there are 4 events, giving 25 ∙ 24 ∙ 23 ∙ 22 = 303,600 ways to choose the officers.

Alternatively, use the permutation formula



In how many ways can a company install 10 different billboards in the 10 locations that the company has rented along a road?

Use the permutation formula with n = 10 and r =10.

There are 3,628,800 ways to install the 10 billboards.


7.6 Example 7 Using the Combinations Formula (page 703)

A college club with 22 members needs to choose 4 members to attend a regional conference. In how many ways can this be done?

Since order is not important, use combinations.

There are 7315 ways to choose the four members.


7.6 Example 8 Using the Combinations Formula (page 703)

Five of the 28 employees of a business are to be selected to serve on this year’s social committee.


There are 98,280 ways to choose the five members.

(a) In how many different ways can the committee members be chosen?


7.6 Example 8 Using the Combinations Formula (cont.)

If last year’s chairperson is included on the committee, then one position is already filled, and the problem becomes choosing four out of the remaining 27 people.

The committee members can be chosen in 17,550 ways if last year’s chairperson must be included.

(b) In how many different ways can the committee members be chosen if last year’s committee chairperson must be included?


7.6 Example 9 Distinguishing Between Permutations and Combinations (page 704)

Should permutations or combinations be used to solve each problem?


(a) In how many ways can 5 oranges be chosen from a bin containing 20 oranges?

Since order is important, use permutations.

(b) A Web site requires each user to select a five-letter password in which no letter may be repeated. How many such passwords are available?


7.6 Example 9 Distinguishing Between Permutations and Combinations (cont.)

Since order is important, use permutations.

(c) In how many ways can 15 employees be assigned to work in 15 cubicles?


(d) In a contest, 3 winners are to be selected at random from among 100 entries. In how many ways can this be done?


Basics of Probability7.7Basic Concepts ▪ Complements and Venn Diagrams ▪ Odds ▪ Union of Two Events ▪ Binomial Probability


7.7 Example 1 Finding Probabilities of Events (page 711)

A single die is rolled. Write each event in set notation and give the probability of the event.

(a) E1: the number showing is a multiple of 3

(b) E2: the number showing is less than 6

For (a) – (d), n(S) = 6.


7.7 Example 1 Finding Probabilities of Events (cont.)

(c) E3: the number showing is 0

(b) E4: the number showing is greater than 0


7.7 Example 2 Using the Complement (page 712)

In the experiment of drawing a card from a well-shuffled deck, find the probabilities of event E, the card is a heart, and of event E′.

There are 52 cards in a deck and 13 of the cards are hearts, so n(S) = 52 and n(E) = 13.


7.7 Example 3 Finding Odds in Favor of an Event (page 713)

A marble is drawn at random from a bowl containing 6 green, 4 blue, and 8 yellow marbles. Find the odds in favor of a yellow marble being drawn.

Let E represent “a yellow marble being drawn.” There are 18 marbles, so n(S) = 18.

The odds in favor of drawing a yellow marble are


7.7 Example 4 Finding Probabilities of Unions (page 714)

One card is drawn from a well-shuffled deck of 52 cards. What is the probability of the following outcomes?

(a) The card is red or a king.

The events “the card is red” and “the card is a king” are not mutually exclusive since it is possible to draw the king of hearts or the king of diamonds, both red cards.


7.7 Example 4 Finding Probabilities of Unions (cont.)

(b) The card is a club or a diamond.

The events “the card is a club” and “the card is a diamond” are mutually exclusive since it is not possible to draw a card that is both a club and a diamond.


7.7 Example 5 Finding Probabilities of Unions (page 714)

Suppose two fair dice are rolled. Find each probability.

(a) The first die shows a 5, or the sum of the two dice is 11 or 12.

Event A has 6 elements, (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), and (5, 6), where (a, b) represents “the first die shows a and the second die shows b.”

There are 36 possible events. Let A represent the event “the first die shows a 5,” and let B represent the event “the sum of the two dice is 11 or 12.”



A and B have one element in common, (5, 6).

Event B has 3 elements, (5, 6), (6, 5), and (6, 6).



(b) The sum of the dots showing is at least 9.

The events represented by “9”, “10”, “11”, or “12” are mutually exclusive.

The “sum of the dots showing is at least a 9” can be written as “the sum of the dots showing is 9, 10, 11, or 12.”



There are 4 ways to sum 9: (6, 3), (5, 4), (4, 5), and (3, 6).

There are 3 ways to sum 10, (6, 4), (5, 5), and (4, 6) .

There are 2 ways to sum 11, (5, 6) and (6, 5) .

There is 1 way to sum 12, (6, 6).


7.7 Example 6 Finding Probabilities in a Binomial Experiment (page 716)

An experiment consists of rolling a die 12 times. Find each probability.

(a) The probability that in exactly 3 of the rolls, the result is a 5.

There are 12 rolls with 3 successes, so n = 12 and r = 3.

The probability p of a 5 on one roll is


7.7 Example 6 Finding Probabilities in a Binomial Experiment (cont.)

(b) The probability that in exactly 8 of the rolls, the result is not a 5.

There are 12 rolls with 8 successes, so n = 12 and r = 8.

The probability p of “not a 5” on one roll is

7 further topics in algebra © 2008 pearson addison-wesley. all rights reserved sections 7.4–7.7

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