# Mathematics

MAT 1301, Liberal Arts Math 1

Course Learning Outcomes for Unit V Upon completion of this unit, students should be able to:

1. Apply mathematical principles used in real-world situations. 1.1 Draw tree diagrams to represent counting situations graphically. 1.2 Apply counting techniques to solve applied problems. 1.3 Apply the theory of permutations and combinations to solve counting problems.

5. Demonstrate counting techniques.

5.1 Count elements in a set systematically. 5.2 Apply the fundamental counting principle. 5.3 Demonstrate how to solve counting problems with special conditions. 5.4 Calculate the number of permutations and combinations of n objects taken r at a time. 5.5 Use factorial notation to represent the number of permutations of a set of objects. 5.6 Create slot diagrams to organize information in counting problems.

Reading Assignment Chapter 12: Counting: Just How Many Are There?

Section 12.1: Introduction to Counting Methods, pp. 606-613

Section 12.2: The Fundamental Counting Principle, pp. 614-621

Section 12.3: Permutations and Combinations, pp. 622-634

Unit Lesson Numbers and counting are some of the first mathematical principles that were introduced to you at a young age. People first learn how to count to ten, then to a hundred, and eventually by even and odd numbers. This unit introduces more principles about counting. In particular, some mathematical properties and formulas that will help in counting larger sets will be learned. 12.1 Systematic Counting: Systematic counting is the act of counting objects in an organized way. For example, if you wanted to list the all the ways that a die could be rolled, you would write “1, 2, 3, 4, 5, 6” because a die has six sides each consisting of a number 1 through 6. Listing these methods in numeric order is a systematic way of finding the answer. The systematic approach will be used to solve the following example. Example: You are selecting from the set of W = {Carrie (U)nderwood, Kelly (C)larkson, Chris (D)aughtry, Fantasia (B)arrino, and Clay (A)iken} This set consists of American Idol contestants who have had successful careers. List all the ways that you can select two singers without repetition; order is not important. For example, BB is not allowed and UC is the same as CU.

UNIT V STUDY GUIDE

Counting

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Solution: A set is a list of items in a list. Set W in the example consists of 5 singers and each singer is denoted by a letter: U, C, D, B, and A. Sets are usually named with a capital letter such as a W. Sets are also usually written with brackets, { }, around the list of items. W = {U, C, D, B, and A} A systematic list will be written to show all the ways in which two singers can be selected. First, pair the first singer (U) with the other singers. This results in the following:

UC, UD, UB, UA Next, pair the second singer (C) with the other singers. The problem states that repetition is not allowed in our list. Therefore, CU will not be included in the list below because UC was previously identified.

CD, CB, CA Next, pair the third singer (D) with the other singers. This will give result in the following:

DB, DA Now we will pair the fourth singer (B) with the other singers. This will result in the following:

BA We cannot pair the last singer (A) with other singers without repeating what was already listed. Thus, the list is complete! There are 10 ways to select two singers from set W. Dice are often used when learning about different counting strategies. The next example uses a systematic counting approach to list the different ways that an event can occur when rolling two dice. Example: Assume that you are rolling two dice: the first one is red and the second is green. Use a systematic listing to determine the number of ways a total of 5 can be rolled. Hint: A total can be rolled two ways: (1, 2) and (2, 1). Solution: A dice has six sides and each side is numbered with either a 1, 2, 3, 4, 5, or 6. To differentiate the two dice, we will assume that one is red and the other green. Dice One: R(red) = {1,2,3,4,5,6} Dice Two: G(green) = {1,2,3,4,5,6} We will use ordered pairs in the form (red, green) to list the number of ways that we can roll a total of 5. First, choose the first number in set R. This number is 1. 1 can only be paired with 4 to make a total of 5.

(1,4)

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Next, choose the second number in set R. This number is 2. The number 2 can only be paired with a 3 to make a total of 5.

(2,3) Next, choose the third number in set R. This number is 3. The numbers 3 can only be paired with 2 to make a total of 5.

(3,2) Next, choose the third number in set R. This number is 4. The number 4 can only be paired with 1 to make a total of 5.

(4,1) Next, choose the fourth number in set R. This number is 5. The number 5 cannot be paired with any number in set G to make a total of 5. The systematic approach ensured that all ordered pairs were identified. Thus, the list is complete. There are 4 ways to list a total of 5 when two dice are rolled. Tree Diagrams A tree diagram illustrates the different ways in which an event can occur and is a graphical approach to counting. Example: Draw a tree diagram that illustrates the different ways to flip a penny, nickel, dime, and quarter. Use this diagram to solve Exercises 5-8 on page 612. In how many ways can exactly two tails be obtained? Solution: To draw the tree, start by writing out the potential outcomes from flipping a penny: heads or tails. Next, branch from those options and list the possible outcomes for flipping a nickel. This process for flipping a dime is continued following by a quarter. Then list the combined sequence of heads and tails for each path of the tree and highlight those that satisfy the condition of getting exactly two tails.

The final answer is 6 ways.

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Example: If a tree diagram was drawn showing how many ways five coins could be flipped, how many branches would it have? Solution: The tree begins at a single point, and then two branches will lead to the results of the first flip. Each of those branches will split into two branches on the second flip for a total of four branches. The third flip will branch from each of those four branches for a total of eight branches. The next flip, flip four, will double the number to sixteen branches. The final flip will double this number one last time for a total of thirty-two ways for five coins being flipped. Example: Assume that a triple-deck ice cream cone with vanilla, strawberry, and chocolate as possible flavors is purchased. The flavors can be repeated or not. Two cones will be considered different if the flavors are the same but occur in different order. How many different flavors are possible? Solution: Since repetition is allowed, each time a choice of flavor is made, there will be three options. Since the order of the flavors do not matter, the multiplication principle is used to solve.

𝐹𝑙𝑎𝑣𝑜𝑟 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 = 3 ∙ 3 ∙ 3

𝐹𝑙𝑎𝑣𝑜𝑟 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 = 27 There are 27 different flavor possibilities. 12.2 The Fundamental Counting Principle: A new car is being purchased, and the individual purchasing the car needs to decide on the features he/she wants to include in the car. The car cannot be built without first deciding on a model, a color, and an audio system. Therefore, the overall task cannot be completed without finishing a series of tasks. The number of ways to build the car could be identified by drawing a tree diagram; however, the fundamental counting principle allows one to identify the number of ways a series of tasks can be completed by a using a simple formula. The Fundamental Counting Principle (FCP) – If you wants to perform a series of tasks, and the first task can be done in a ways, the second can be done in b ways, the third can be done in c ways, and so on, then all tasks can be done in a × b × c × … ways. Assume that you can select among 3 models, 5 colors, and 2 audio systems when building a car. By using the FCP formula, you know that there are 3 × 5 × 2 = 30 ways to build your new car. Example: The Equestrian Club has eight members. If the club wants to select a president, vice president, and treasurer (all of whom must be different), in how many ways can this be done? Solution: By applying the fundamental counting principle, we can see that there are eight ways that the first position can be filled (one of the eight members). Once that position is filled, there are seven ways that the second position can be filled (one of the seven remaining members). Moving on to the third position, there are only six

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ways that it can be filled (one of the six remaining members). Therefore, those possibilities can be multiplied to obtain the result:

8 ∙ 7 ∙ 6 = 𝟑𝟑𝟔 different ways Example: The early bird special at TGI Friday’s features an appetizer, soup or salad, entrée, and dessert. If there are 5 appetizers, 6 choices for soup or salad, 13 entrées, and 4 desserts, how many different meals are possible? (It is assumed that one makes a selection from each category.) Solution: To solve this problem, the number of ways that each food category may be selected needs to be determined. There are 5 appetizers, 6 choices for soup or salad, 13 entrées, and 4 desserts. Therefore, the fundamental counting principle can be used, and all choices can be multiplied together to find the result.

5 ∙ 6 ∙ 13 ∙ 4 = 1,560 There are 1,560 different ways to choose a meal. 12.3 Permutations and Combinations: Permutations and combinations are mathematical terms that are used when solving detailed counting problems. Factorial Notation A factorial is a mathematical operation used when computing permutations and combinations. It is denoted by an exclamation mark (!). For example, n! means that the factorial operation will be applied to n and is read “n factorial”. Definition – If n is a counting number, the symbol n! stands for the product n∙(n-1)∙(n-2)∙(n-3)∙∙∙∙2∙1. One defines 0! = 1. Example: Solve: 9! Solution: The definition of a factorial will be used to solve. n! = n∙(n-1)∙(n-2)∙(n-3)∙∙∙∙2∙1 = 9∙(9-1)∙(9-2)∙(9-3)∙∙∙∙2∙1 = 9∙8∙7∙6∙5∙4∙3∙2∙1 Note: Keep subtracting until 1 is obtained as the last digit in the sequence. = 563,760 Example: Solve: 10!

7!

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Solution: 10!

7!

= 10∙9∙8∙7∙6∙5∙4∙3∙2∙1

7∙6∙5∙4∙3∙2∙1 Write the definition of each factorial.

= 10∙9∙8∙7∙6∙5∙4∙3∙2∙1

7∙6∙5∙4∙3∙2∙1 Cancel like factors on the top and bottom of the fraction.

= 10 ∙ 9 ∙ 8 Multiply. = 𝟕𝟐𝟎 Example: Solve: 10!

7! 3!

Solution: 10!

7! 3!

= 10∙9∙8∙7∙6∙5∙4∙3∙2∙1

(7∙6∙5∙4∙3∙2∙1)(3∙2∙1) Write the definition of each factorial.

= 10∙9∙8∙7∙6∙5∙4∙3∙2∙1

(7∙6∙5∙4∙3∙2∙1)(3∙2∙1) Cancel like factors on the top and bottom of the fraction.

= 10∙9∙8

3∙2∙1 Multiply.

= 720

6 Divide.

= 𝟏𝟐𝟎 Permutations A permutation is the act of rearranging a set of numbers or items in an order or sequence without repetition. For example, suppose you have the following set of letters:

S = {a, b, c} This set has 6 permutations because the letters can be arranged 6 different ways without repetition. These ways are illustrated below:

abc, acb, bac, bca, cba, cab

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The mathematical notation used to express permutations is denoted by 𝑃(𝑛, 𝑟). This is read as “the number of permutations of n objects taken r at a time.” The figure below illustrates this and may be found on page 623 of the textbook.

There were 3 letters selected from set S in the example above. All 3 letters were chosen to make our permutations. Therefore, it was found that 𝑃(𝑛, 𝑟) = 𝑃(3, 3) = 6. The knowledge of factorials can be used to compute permutations using a formula. Formula for computing 𝑃(𝑛, 𝑟):

𝑃(𝑛, 𝑟) = 𝑛!

(𝑛 − 𝑟)!

Example: Find the permutation. Eight objects taken three at a time. Solution: First, identify n and r. There are eight objects, so n = 8. Three objects are taken at a time, so r = 3. Next, solve 𝑃(𝑛, 𝑟) = 𝑃(8,3). 𝑃(8,3)

= 8!

(8−3)! Plug in the values for n and r in the permutation formula.

= 8!

5! Subtract the values in the parentheses.

= 8∙7∙6∙5∙4∙3∙2∙1

5∙4∙3∙2∙1 Write the definition of each factorial.

= 8∙7∙6∙5∙4∙3∙2∙1

5∙4∙3∙2∙1 Cancel like factors on the top and bottom of the fraction.

= 8 ∙ 7 ∙ 6 Multiply. = 𝟑𝟑𝟔 Example:

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Solve. 𝑃(10,3) Solution: 𝑃(10,3)

= 10!

(10−3)! Plug in the values for n and r in the permutation formula.

= 10!

7! Subtract the values in the parentheses.

= 10∙9∙8∙7∙6∙5∙4∙3∙2∙1

7∙6∙5∙4∙3∙2∙1 Write the definition of each factorial.

= 10∙9∙8∙7∙6∙5∙4∙3∙2∙1

7∙6∙5∙4∙3∙2∙1 Cancel like factors on the top and bottom of the fraction

= 10 ∙ 9 ∙ 8 Multiply. = 𝟕𝟐𝟎 Identifying Permutations in Word Problems The following permutation problems will help to identify when permutations should be calculated. Remember that a permutation will be calculated when order matters, and items in a list cannot be repeated. Example: On a biology quiz, a student must match eight terms with their definitions. Assume that the same term cannot be used twice. Write the formula that will be used. Solving the problem is not required. Solution: The key to this problem is to know that the order of matching the terms with their definitions matters, thus this is a permutation. Therefore, it is denoted as P(8,8). Example: A password for a computer consists of three different letters of the alphabet followed by four different digits from 0 to 9. How many different passwords are possible? Solution: In this situation the order matters, so this is a permutation. Three letters will be chosen from a possible 26, and 4 digits from a possible 10. The two results will then be multiplied by each other for the final result.

𝑃(26,3) = 26!

(26 − 3)! =

26!

23! =

26 ∙ 25 ∙ 24 ∙ 23!

23! = 26 ∙ 25 ∙ 24 = 15,600

𝑃(10,4) = 10!

(10 − 4)! =

10!

6! =

10 ∙ 9 ∙ 8 ∙ 7 ∙ 6!

6! = 10 ∙ 9 ∙ 8 ∙ 7 = 5040

Therefore, multiplying both results together results in the total number of passwords possible: 15,600 ∙ 5040 = 𝟕𝟖, 𝟔𝟐𝟒, 𝟎𝟎𝟎 possible passwords.

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Combinations Combinations can be described as selecting items from a group when the order of the selection does not matter. In other words, group (abc) would by the same as (bca), because the order does not matter. The mathematical notation used to express combinations is denoted by 𝐶(𝑛, 𝑟). This is expressed as follows: “We are forming a combination of n objects taken r at a time.” Formula for computing 𝐶(𝑛, 𝑟):

𝐶(𝑛, 𝑟) = 𝑃(𝑛, 𝑟)

𝑟! =

𝑛!

𝑟! ∙ (𝑛 − 𝑟)!

Example: Find the number of combinations. Eight objects taken three at a time. Solution: C(8,3)

= 8!

3!(8−3)! Plug in the values for n and r in the combination formula.

= 8!

3!5! Subtract the values in the parentheses.

= 8∙7∙6∙5∙4∙3∙2∙1

(3∙2∙1)(5∙4∙3∙2∙1) Write the definition of each factorial.

= 8∙7∙6∙5∙4∙3∙2∙1

(3∙2∙1)(5∙4∙3∙2∙1) Cancel like factors on the top and bottom of the fraction.

= 8∙7∙6

3∙2∙1 Multiply.

= 336

6 Divide.

= 𝟓𝟔 combinations Example: Solve. 𝐶(6,2) Solution: 𝐶(6,2)

= 6!

2!(6−2)! Plug in the values for n and r in the combination formula.

= 6!

2!4! Subtract the values in the parentheses.

= 6∙5∙4∙3∙2∙1

(2∙1)(4∙3∙2∙1) Write the definition of each factorial.

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= 6∙5∙4∙3∙2∙1

(2∙1)(4∙3∙2∙1) Cancel like factors on the top and bottom of the fraction.

= 6∙5

2∙1 Multiply.

= 30

2 Divide.

= 𝟏𝟓 Identifying Combinations in Word Problems The following combination problems will help you identify when combinations should be calculated. Remember that a combination will be calculated when the order of the items does not matter. Example: A committee is to be formed consisting of 5 men and 4 women. If the membership is to be chosen from 12 men and 10 women, how many different committees are possible? Solution: This is a combination problem so the formula below will be used:

𝐶(𝑛, 𝑟) = 𝑛!

𝑟! (𝑛 − 𝑟)!

Where “n” is the number of elements taken “r” at a time. Step 1: Find the number of ways to choose the men. Out of the 12 possible men, 5 of them should be chosen.

𝐶(12,5) = 12!

(12 − 5)! 5! =

12 ∙ 11 ∙ 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

(7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1)(5 ∙ 4 ∙ 3 ∙ 2 ∙ 1) =

12 ∙ 11 ∙ 10 ∙ 9 ∙ 8

5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =

95040

120 = 792 𝑤𝑎𝑦𝑠

Step 2: Find the number of ways to choose the women. Out of the 10 possible women, 4 of them should be chosen.

𝐶(10,4) = 10!

(10 − 4)! 4! =

10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

(6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1)(4 ∙ 3 ∙ 2 ∙ 1) =

10 ∙ 9 ∙ 8 ∙ 7

4 ∙ 3 ∙ 2 ∙ 1 =

5040

24 = 210 𝑤𝑎𝑦𝑠

Step 3: Multiply the number of ways to choose 5 men and 4 women.

𝐶(12,5) ∙ 𝐶(10,4) = 792 ∙ 210 = 166,320 Thus, there are 166,320 ways to choose 5 men from 12 men and to choose 4 women from 10 women. Example: A pet store has 6 fluffy bunnies and 9 hamsters. In how many ways can 3 animals be selected if there can be at most 1 fluffy bunny? Solution: To answer this, all the ways that one or less fluffy bunnies can be chosen need to be looked at. 1. If one chooses 0 fluffy bunnies, then 3 hamsters are chosen. 2. If one chooses 1 fluffy bunny, then 2 hamsters are chosen.

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So for each one, the formula for combinations should be used:

𝐶(𝑛, 𝑟) = 𝑛!

𝑟! (𝑛 − 𝑟)!

Where “n” is the number of elements taken “r” at a time. 1. 0 fluffy bunnies are chosen out of 6 fluffy bunnies, and 3 hamsters are chosen out of 9 hamsters.

𝐶(6,0) ∙ 𝐶(9,3) = 6!

(6 − 0)! 0! ∙

9!

(9 − 3)! 3! =

6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

(6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1)(1) ∙

9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

(6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1)(3 ∙ 2 ∙ 1)

= 1 ∙ 84 = 84 2. 1 fluffy bunny is chosen out of 6 fluffy bunnies, and 2 hamsters are chosen out of 9 hamsters.

𝐶(6,1) ∙ 𝐶(9,2) = 6!

(6 − 1)! 1! ∙

9!

(9 − 2)! 2! =

6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

(5 ∙ 4 ∙ 3 ∙ 2 ∙ 1)(1) ∙

9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

(7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1)(2 ∙ 1)

= 6 ∙ 36 = 216 Now, the possible combinations need to be added together.

84 + 216 = 𝟑𝟎𝟎 𝒘𝒂𝒚𝒔

Reference

Pirnot, T. L. (2014). Mathematics all around (5th ed.). Boston, MA: Pearson.