Part 1 Module 4 The Fundamental Counting Principle

EXAMPLE 1.4.1

Plato is going to choose a three-course meal at his favorite restaurant. He must choose one item from each of the following three categories.

First course

: Tofu Soup (

TS

); Seaweed Salad (

SS

)

Second course

: Steamed Tofu (

ST

); Baked Tofu (

BT

); Fried Tofu (

FT

);

Third course

: Tofu Cake (

TC

); Tofu Pie (

TP

); Seaweed Delight (

SD

) How many different three course meals are possible? A. 12 B. 18 C. 8 D. None of these P1M4 1

Part 1 Module 4 The Fundamental Counting Principle

We will list every possible 3-course meal: 1. TS-ST-TC 2. TS-ST-TP 3. TS-ST-SD 4. TS-BT-TC 5. TS-BT-TP 6. TS-BT-SD 7. TS-FT-TC 8. TS-FT-TP 9. TS-FT-SD 10. SS-ST-TC 11. SS-ST-TP 12. SS-ST-SD 13. SS-BT-TC 14. SS-BT-TP 15. SS-BT-SD 16. SS-FT-TC 17. SS-FT-TP 18. SS-FT-SD There are 18 different 3-course meals.

P1M4 2

The Fundamental Counting Principle

You probably noticed that there is a more economical way to answer that question.

Choosing a three-course meal requires three independent decisions: 1. Choose first course item (2 options).

2. Choose second course item (3 options) 3. Chooose third course item (3 options)

2x3x3 = 18 different three-course meals.

P1M4 3

The Fundamental Counting Principle

This illustrates the

Fundamental Counting Principle

, which describes a technique for determining the number of different outcomes in certain complex processes: Step 1: Analytically break down the complex process into a number of distinct stages or

decisions

; Step 2: Determine the number of options for each decision identified in Step 1; Step 3: Multiply the numbers from Step 2.

P1M4 4

Why it works

The

Fundamental Counting Principle

works when we have an orderly decision process that has an underlying tree structure. We are counting the number a branch-tips at the end of the tree.

P1M4 5

Why it works

FIRST COURSE TS SECOND COURSE ST FT BT THIRD COURSE TC TP SD TC TP SD TC TP SD TS-ST-TC TS-ST-TP TS-ST-SD TS-FT-TC TS-FT-TP TS-FT-SD TS-BT-TC TS-BT-TP TS-BT-SD SS ST FT BT TC TP SD SS-ST-TC SS-ST-TP TC TP SD SS-ST-SD SS-FT-TC SS-FT-TP TC TP SD SS-FT-SD SS-BT-TC SS-BT-TP SS-BT-SD

P1M4 6

Fundamental Counting Principle The simplest Fundamental Counting Principle problems are those in which we are presented with a menu, and the situation dictates that we must choose one item from each category on the menu.

P1M4 7

EXAMPLE 1.4.12

Gomer is considering the purchase of a new super-cheap sport/utility vehicle, the Skuzuzi Kamikaze. He must choose a vehicle, taking into account the following options: i.

Transmission:

4-speed standard transmission, 5-speed standard transmission, or automatic transmission; ii.

Bumper:

steel bumpers, vinyl bumpers or 2x4 boards bolted to the front and back; iii.

Top:

hard-top, vinyl top convertible, or chicken wire stapled over the roll bar; iv.

Funerary accessory:

urn.

complementary funeral wreath or cremation 1. How many different vehicle option packages are possible? 2. How many packages are possible if he already knows that he will order the chicken wire and can ’ t order the steel bumpers?

P1M4 8

Solution

1.4.12 - 1

1. How many different vehicle option packages are possible?

Gomer must make four independent decisions.

i.

Transmission: 3 options

(4-speed standard transmission, 5-speed standard transmission, or automatic transmission}; ii.

Bumper: 3 options

(steel bumpers, vinyl bumpers or 2x4 boards bolted to the front and back); iii.

Top: 3 options

(hard-top, vinyl top convertible, or chicken wire stapled over the roll bar); iv.

Funerary accessory: 2 options

cremation urn) (complementary funeral wreath or According to the Fundamental Counting Principal, the number of outcomes for this decision process is (3)(3)(3)(2) = 54 P1M4 9

Solution

1.4.12 - 2

2. How many packages are possible if he already knows that he will order the chicken wire and can

’

t order the steel bumpers?

Again, Gomer must make four independent decisions. We must take into account the effects of the predetermined conditions.

i.

Transmission: 3 options

(4-speed standard transmission, 5-speed standard transmission, or automatic transmission}; ii.

Bumper: 2 options

and back); ( vinyl bumpers or 2x4 boards bolted to the front iii.

Top: 1 option

(chicken wire stapled over the roll bar); iv.

Funerary accessory: 2 options

cremation urn) (complementary funeral wreath or According to the Fundamental Counting Principal, the number of outcomes for this decision process is (3)(2)(1)(2) = 12 P1M4 10

Exercise

Prior to the start of a polo match, the captains of the two teams gather in the country club lounge. One team has 4 captains, and the other team has 3 captains.

If each captain of the first team shakes hands with each captain of the second team, how many handshakes occur?

P1M4 11

Exercise

A quiz consists of 3 true/false questions followed by 4 multiple choice questions, with 4 options for each multiple-choice question.

In how many ways is it possible to answer the 7 questions?

A. 264 B. 507 C. 2424 D. None of these. The number of ways to answer the 3 true/false questions is 2 3 = 8.

The number of ways to answer the 4 multiple-choice questions is 4 4 = 256.

Thus, the number of ways to answer all 7 questions is (8)(256) = 2048.

The correct answer is 'D. '

P1M4 12

EXAMPLE 1.4.6

There are 5 guys (including Gomer) on Gomer's bowling team. After their game they will each choose one of the following:

Dr. Pauper, Dr. Thunder, or Dr. Special

. How many outcomes are possible?

P1M4 13

SOLUTION 1.4.6

Since there are five guys, each of whom must make a decision, this counting process involves five independent decisions. Call the five guys Al, Bill, Carl, Doug and Gomer.

Al has 3 options (he can choose

Dr. Pauper, Dr. Thunder, or Dr. Special

).

Bill has 3 options.

Carl has 3 options.

Doug has 3 options.

Gomer has 3 options.

According to the Fundamental Counting Principle, the number of outcomes is (3)(3)(3)(3)(3) = 243 P1M4 14

EXAMPLE 1.4.6 - 2

Again, there are 5 guys (including Gomer) on Gomer's bowling team. After the game they will each choose one of the following:

Dr. Pauper, Dr. Thunder, or Dr. Special

. However, Bill and Doug are having a spat, so they never agree on anything.

How many outcomes are possible, assuming that Bill and Doug will not order the same product?

P1M4 15

SOLUTION 1.4.6 - 2

This counting process involves five decisions, but two of the decisions (Bill ’ s and Doug Bill, Carl, Doug and Gomer.

’ s) influence one another. Call the five guys Al, Al has 3 options (he can choose

Dr. Pauper, Dr. Thunder, or Dr. Special

).

Bill has 3 options.

Carl has 3 options.

Doug has only 2 options (he cannot choose whatever product was chosen by Bill).

Gomer has 3 options.

According to the Fundamental Counting Principle, the number of outcomes is (3)(3)(3)(2)(3) = 162 P1M4 16

P1M4 17

72 Dr. Pepper Clones

If each of the five guys in our previous example drank one of the 72 displayed Dr. Pepper clones at the end of their bowling game, how many possible outcomes of choices are there?

Solution: Each of the five guys has 72 choices, meaning there are 5 independent decisions being made with 72 options for each decision. According to the Fundamental counting Principle, the number of outcomes is (72)(72)(72)(72)(72) = 1,934,917,632 P1M4 18

Exercise

The dial on a combination lock has numbers ranging from 1 to 30. The “ combination ” that opens the lock is a sequence of three numbers.

How many different combinations are possible, assuming that the combination may have repeated numbers, but the same number will not appear twice consecutively?

For example, 29-15-8, 7-13-22, 14-2-14, 8-29-15 are four different possible combinations, but 5-5-12 and 3-16-16 are not acceptable.

P1M4 19

Solution

To create a three-number sequence, such as 15-3-18, requires three decisions: i. Choose first number: 30 options ii. Choose second number: 29 options (whichever one of the thirty numbers was chosen for the first number is not available, because the same number cannot be used twice in a row).

iii.Choose third number: 29 options (whichever one of the thirty numbers was chosen for the second number is not available, because the same number cannot be used twice in a row).

30x29x29 = 25,230 different outcomes (choice C).

P1M4 20

Example 1.4.11 #1

How many different 4-digit numbers can be formed using the following digits? {0, 2, 3, 5, 8} Note: the first digit cannot be 0, or else the number would be a 3-digit number.

P1M4 21

Solution 1.4.11 #1

How many different 4-digit numbers can be formed using the following digits? {0, 2, 3, 5, 8} In order to form a 4-digit number we must make four independent decisions: i.

ii.

Choose first digit: 8).

4 options

(the first digit could be 2, 3, 5, or 8).

Choose second digit:

5 options

(the second digit could be 0, 2, 3, 5, or iii.

iv.

Choose third digit:

5 options

Choose fourth digit: (the third digit could be 0, 2, 3, 5, or 8).

5 options

(the fourth digit could be 0, 2, 3, 5, or 8).

According to the Fundamental Counting Principle the number of outcomes is(4)(5)(5)(5) = 500. There are 500 possible 4-digit numbers.

P1M4 22

Example 1.4.11 #2

How many different 4-digit numbers can be formed using the following digits, assuming that the 4-digit number is a multiple of five? {0, 2, 3, 5, 8} A. 100 B. 4552 C. 4551 D. 200 E. None of these P1M4 23

Solution 1.4.11 #2

How many different 4-digit numbers can be formed using the following digits, assuming that the 4-digit number is a multiple of five? {0, 2, 3, 5, 8} In order to form one of these numbers, again we need to make 4 decisions. However, in this case the last digit must be either 0 or 5 (in order for the number to be a multiple of 5).

i. Choose first digit: 4 options (the first digit could be 2, 3, 5, or 8).

ii. Choose second digit: 5 options (the second digit could be 0, 2, 3, 5, or 8).

iii. Choose third digit: 5 options (the third digit could be 0, 2, 3, 5, or 8).

iv. Choose fourth digit: 2 options (the fourth digit must be 0 or 5).

According to the Fundamental Counting Principle the number of outcomes is (4)(5)(5)(2) = 200. There are 200 possible 4-digit multiples of 5.

P1M4 24

Example

Gomer is going to order a frozen tofu cone from

I Definitely Believe It's Tofu.

The following toppings are available: 1. carob chips 2. frosted alfala sprouts 3. seaweed sprinkles 4. rolled oats 5. rose hips He may choose all, some or none of these toppings. How many topping combinations are possible? A. 5 B. 10 C. 25 D. 32 E. 120 P1M4 25

What are we trying to count?

Five toppings: CC, FAS, SS, RO, RH Here are several different combinations of items.

1. RO, SS 2. FAS 3. RH, CC, FAS 4. SS, CC 5. CC, FAS, RO, RH 6. (blank: this means we didn ’ t choose any of the toppings) 7. CC, FAS, SS, RO, RH 8. SS How many of these are possible?

P1M4 26

What are we trying to count?

We have a set of five toppings: {CC, FAS, SS, RO, RH} Here are several different combinations of items.

1. {RO, SS} 2. {FAS} 3. {RH, CC, FAS} 4. {SS, CC} 5. {CC, FAS, RO, RH} 6. { } (this means we didn ’ t choose any of the toppings) 7. {CC, FAS, SS, RO, RH} 8. {SS} Each combination of toppings is a subset of the five-topping set, so we are just counting the number of subsets in a set with five elements, which is 2 5 P1M4 = 32.

27

Alternative solution, using FCP

We have a set of five toppings: {CC, FAS, SS, RO, RH} We have shown that the number of different combinations of toppings is 32, because each combination of toppings is a subset of this set, and a set with 5 elements has 32 subsets.

We can also get this answer by using the FCP. When we select a combination of toppings we are actually making five “ yes/no ” decisions: 1. carob chips: 2 options ( “ yes ” 2. frosted alfalfa sprouts: 2 options ( 3. seaweed sprinkles: 2 options ( or “ no ” ) “ “ yes yes ” ” or “ no ” ) or “ no ” ) 4. rolled oats: 2 options ( “ yes ” 5. rose hips: 2 options ( “ yes ” or “ no ” or “ no ” ) ) According to the FCP, the number of outcomes is 2  2  2  2  2 = 32 P1M4 28

“

All, some, or none

” If a counting problem involves an “ all, some, or none ” situation, then the number of outcomes will always be a power of 2 (such as 2, 4, 8, 16, 32, 64, 128, 256 and so on).

This is because the situation involves a series of two-way (

“

yes or no

”

) decisions, so the Fundamental Counting Principle will have us multiplying a series of twos.

This also due to the fact that such a problem is asking for the number of subsets in a particular set.

P1M4 29

Exercise

Homerina is having a birthday party for her pet wolverine, John. John has a list of 9 gifts that he would like to receive: a duck, a hamster, a puppy, a mouse, a goldfish, a frog, a toad, a chicken, and a claw sharpener.

How many combinations of gifts are possible, assuming that Homerina may buy all, or some, or none of those items?

A. 81 B. 1280 C. 512 D. 18 E. None of these P1M4 30

Solution

For each of the nine gifts, Homerina faces a “ yes/no ” decision, so this is an FCP problem featuring nine independent decisions: 1. duck: 2 options ( “ yes ” or “ no ” ) 2. hamster: 2 options 3. puppy: 2 options 4. mouse: 2 options 5. goldfish: 2 options 6. frog: 2 options 7. toad: 2 options 8. chicken: 2 options 9. claw sharpener: 2 options According to the FCP, the number of outcomes is 2  2  2  2  2  2  2  2  31

Exercise

Hjalmar, Gomer, Plato, Euclid, Socrates, Aristotle, Hjalmarina and Gomerina form the board of directors of the Lawyer and Poodle Admirers Club. They will choose from amongst themselves a Chairperson, Secretary, and Treasurer. No person will hold more than one position. How many different outcomes are possible?

A. 336 B. 24 C. 512 D. 21 P1M4 32

Solution

Choosing a Chairperson, Secretary and Treasurer from among these 8 people requires us to make three decisions. NOTICE these three decisions are not independent. For instance, the choice we make when we select the chairperson affects which options are available when we go to choose the Secretary, since the person selected to be Chairperson cannot also be selected to be Secretary.

i. Choose Chairperson: 8 options; ii. Choose Secretary: 7 options (one person has already been chosen to be Chairperson); iii. Choose Treasurer: 6 options (two people have already been chosen to be Chairperson and Secretary, respectively).

According to the Fundamental Counting Principle the number of outcomes is: (8)(7)(6) = 336.

P1M4 33

Exercise

Erasmus is trying to guess the combination to his combination lock. The "combination" is a sequence of three numbers, where the numbers range from 1 to 12, with

no numbers repeated

. How many different "combinations" are possible if he knows that the last number in the combination is either 1 or 11?

A. 264 D. 288 B. 1320 E. 180 C. 220 P1M4 34

Solution

Choosing a three-number sequence having no repeated numbers requires that we make three dependent decisions. One of these decisions, however, has a special condition attached to it (the third number must be either 1 or 11).

When using the Fundamental Counting Principle in a situation involving dependent decisions, if one decision has a special condition, that decision must be treated first, because the special condition overrides the other decisions

. For example, that fact that the third number must be 1 or 11 means that it is impossible for the sequence to simultaneously have 1 for the first number and 11 for the second number (since then there would be nothing left for the third number).

Three dependent decisions: 1. Choose

third

number (two options); 2. Choose first number (11 options); 3. Choose second number (10 options).

According to the Fundamental Counting Principle the number of possible outcomes is (2)(11)(10) = 220.

P1M4 35

Dependent decisions

If a Fundamental Counting Principle problem involves

dependent decisions

, and one decision involves a special condition,

the decision with the special condition takes priority over the others

.

P1M4 36

Example

The Egotists' Club has 6 members: A, B, C, D, E, and F. They are going to line up, from left to right, for a group photo. After lining up in alphabetical order (ABCDEF), Mr. F complains that he is always last whenever they do things alphabetically, so they agree to line up in reverse order (FEDCBA) and take another picture. Then Ms. D complains that she's always stuck next to Mr. C, and that she never gets to be first in line. Finally, in order to avoid bruised egos, they all agree to take pictures for every possible left-to-right line-up of the six people. How many different photos must be taken? A. 14 C. 823,543 B. 720 D. None of these P1M4 37

Solution

To form a six-person line-up requires six dependent decisions: Choose first (leftmost) person: 6 options Choose second person: 5 options Choose third person: 4 options Choose fourth person: 3 options Choose fifth person: 2 options Choose last (rightmost) person: 1 option 6  5  4  3  2 

permutations

1 = 720 different

arrangements

on six people in a row.

or P1M4 38

Similar situations

Suppose we had the same situation, but with 8 people instead of six. Then the number of ways to arrange them in a row would be 8  7  6  5  4  3  2  1 = 40,320 Likewise, if there were ten people, the number of arrangements would be 10  9  8  7  6  5  4  3  2  1 = 3,628,800 and so on P1M4 39

Factorials

Numbers like 6  8  10  5  7  9  4  6  8  3  5  7  2  4  6  1 3  5  2  4  1 3  2  1 are called

factorials.

P1M4 40

Factorials

6  5  4  denoted 6!

3  2  1 is called “ 6 factorial ” 8  7  6  denoted 8!

5  4  3  2  1 is called “ 8 factorial ” 10  9  8  factorial ” 7  6  denoted 10!

5  4  3  2  1 is called “ 10 P1M4 41

“

n factorial

” If

n

is a positive integer, then

n factorial

, denoted

n!

, is the number

n

multiplied by all the smaller positive integers.

n! = n



(n –1)



(n –2)



…



3



2



1 n! is the number of ways to arrange n objects.

Also,

0! = 1

P1M4 42

Exercise

Macbeth is trying to guess the password for Gomerina's email account. He knows that the password consists of 4 letters chosen from this set: {g,o,m,e,r,i,n,a}.

How many passwords are possible, if a password does not contain repeated letters and the third letter is a vowel?

A. 32 B. 256 C. 840 D. 1344 E. None of these P1M4 43

Solution

Macbeth is trying to guess the password for Gomerina's email account. He knows that the password consists of 4 letters chosen from this set: {g,o,m,e,r,i,n,a}.

How many passwords are possible, if a password does not contain repeated letters and the third letter is a vowel?

In order to guess a password Macbeth has to make four dependent decisions. Since the choice of the third letter affects the other choices, that decision must be made first. There are four options for the third letter, and then there are seven options for the first letter, six options for the second letter, and five options for the last letter.

According to the Fundamental Counting Principle, the number of possible outcomes is (4)(7)(6)(5) = 840.

P1M4 44

“

All, some, or none

” If a counting problem involves an “ all, some, or none ” situation, then the number of outcomes will always be a power of 2.

This is because the situation involves a series of two-way ( “ yes or no ” ) decisions.

This also due to the fact that such a problem is asking for the number of subsets in a particular set.

P1M4 45