Part 1 Module 4 The Fundamental Counting Principle

1. 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? Part 1 Module 4The Fundamental Counting Principle A. 12 B. 18 C. 8 D. None of these

2. We will list every possible 3-course meal: Part 1 Module 4The Fundamental Counting Principle There are 18 different 3-course meals.

3. 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. The Fundamental Counting Principle

4. 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. The Fundamental Counting Principle

5. 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. Why it works

6. Why it works

7. 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. Fundamental Counting Principle

8. 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: complementary funeral wreath or cremation urn. 1. How many different vehicle option packages are possible? A. 54 B. 11 C. 81 D. None of these 2. How many packages are possible if he already knows that he will order the chicken wire and can’t order the steel bumpers? EXAMPLE 1.4.12

9. 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 (complementary funeral wreath or cremation urn) According to the Fundamental Counting Principal, the number of outcomes for this decision process is (3)(3)(3)(2) = 54 Solution 1.4.12 - 1

10. 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 ( vinyl bumpers or 2x4 boards bolted to the front and back); iii. Top:1 option (chicken wire stapled over the roll bar); iv. Funerary accessory: 2 options (complementary funeral wreath or cremation urn) According to the Fundamental Counting Principal, the number of outcomes for this decision process is (3)(2)(1)(2) = 12 Solution 1.4.12 - 2

11. There are 5 guys (including Gomer) on Gomer's bowling team. After the beer frame they will each choose one of the following: Scud, Scud Lite, or Scud Ice. How many outcomes are possible? A. 60 B. 125 C. 15 D. 243 E. None of these EXAMPLE 1.4.6

12. 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 Scud, Scud Lite, or Scud Ice). 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) = 343 SOLUTION 1.4.6

13. Again, there are 5 guys (including Gomer) on Gomer's bowling team. After the beer frame they will each choose one of the following: Scud, Scud Lite, or Scud Ice. 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? A. 27 B. 108 C. 81 D. None of these EXAMPLE 1.4.6 - 2

14. This counting process involves five decisions, but two of the decisions (Bill’s and Doug’s) influence one another. Call the five guys Al, Bill, Carl, Doug and Gomer. Al has 3 options (he can choose Scud, Scud Lite, or Scud Ice). 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 SOLUTION 1.4.6 - 2

15. 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. A. 27,000 B. 24,360 C. 25,230 D. None of these

16. 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).

17. 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.

18. Solution 1.4.11 #11 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: Choose first digit: 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 8). Choose third digit: 5 options (the third digit could be 0, 2, 3, 5, or 8). Choose fourth digit: 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.

19. 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

20. 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.

21. 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? 5 B. 10 C. 25 D. 32 E. 120

22. 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?

23. 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 25 = 32.

24. 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” or “no”) 2. frosted alfalfa sprouts: 2 options (“yes” or “no”) 3. seaweed sprinkles: 2 options (“yes” or “no”) 4. rolled oats: 2 options (“yes” or “no”) 5. rose hips: 2 options (“yes” or “no”) According to the FCP, the number of outcomes is 2  2  2  2  2 = 32

25. “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.

26. 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

27. 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  2 = 512

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. 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.

29. 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

30. Solution Choosing a Chairperson, Secretary and Treasurer from among these 8 people requires us to make three decisions. However (unlike in all of the previous examples) 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.

31. 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 B. 1320 C. 220 D. 288 E. 180

32. 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.

33. 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.

34. 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? 14 B. 720 C. 823,543 D. None of these

35. 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  1 = 720 different arrangements or permutations on six people in a row.

36. 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

37. Factorials Numbers like 6  5  4  3  2  1 8  7  6  5  4  3  2  1 10  9  8  7  6  5  4  3  2  1 are called factorials.

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

39. “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

40. 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

41. 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.