Permutations

1. Permutations Counting Techniques, Part 1

2. Permutations Recall the notion of calculating the number of ways in which to arrange a given number of items. One tool for accomplishing this task is the Multiplication Principle.

3. Permutations One way to state the multiplication principle is that to determine the possible number of outcomes in a given situation, multiply together the number of options available every time you make a choice. For example, say you want to order a sundae with 10 flavors of ice cream and 3 different toppings available.

4. Permutations For example, say you want to order a sundae with 10 flavors of ice cream and 3 different toppings available. The total possible number of sundaes using one scoop of ice cream and one topping is 10*3, or 30.

5. Permutations Another way to state the multiplication principle is this: If event A can take place in m ways, and event B can take place in n ways, then event A and event B can take place in m*n ways. Example: Consider a game where the player flips a coin and rolls a die.

6. Permutations There are 2 possible outcomes for the coin (Heads or Tails) and 6 possible outcomes for the die (1-6), so there are 2*6, or 12 possible outcomes for the two events considered together. Example: Consider a game where the player flips a coin and rolls a die. How many outcomes are possible for flipping a coin and rolling a die?

7. Permutations Here’s one statement of the addition principle: If event A can occur in m ways and event B can occur in n ways, then event A or event B can occur in m + n ways. In some instances, determining the possible number of outcomes requires the use of the Addition Principle.

8. Permutations A school club consists of 14 Juniors and 18 Seniors. In how many ways could we select a committee of 2 students if one must be a Junior and one must be a Senior? We have 14 options for the Junior, and 18 options for the Senior, so we have a total of 14*18, or 252 possibilities. Comparing the multiplication and addition principles:

9. Permutations A school club consists of 14 Juniors and 18 Seniors. How many options do we have for a representative if either a Junior or Senior can fill the position? We have 14 options for a Junior, and 18 options for a Senior, so we have a total of 14 + 18, or 32 possibilities. Comparing the multiplication and addition principles:

10. Permutations Recall that in some cases we needed to modify the multiplication principle—namely when order does not matter. For example, when picking two numbers from a list of 10, we have (10*9)/2, or 45 possibilities. Comparing the multiplication and addition principles:

11. Permutations We needed to divide by 2 to take into account the fact that we had counted every pair of numbers twice. I.e., we counted (1, 2) as well as (2, 1) even though they are the same pair of numbers, and should only count once. Comparing the multiplication and addition principles:

12. Permutations Likewise, we sometimes need to modify the addition principle. Comparing the multiplication and addition principles:

13. Permutations Consider a committee that requires either a soccer player or a track athlete. For this example, say the athletes are in a school where they cannot compete in both soccer and track, due to the two sports being in the same season. If there are 40 soccer players and 60 track athletes, how many possibilities are there to select a representative?

14. Permutations Track Soccer 60 40 Because an athlete cannot compete in both sports in this case: We have 40 + 60, or 100 possibilities.

15. Permutations Now consider a committee that requires either a soccer player or a band member. If there are 40 soccer players and 80 band members, how many possibilities are there to select a representative?

16. Permutations Band Soccer 80 40 In this case, a student could be a member of both groups: Because of the overlap between the two groups, we cannot simply add 40 + 80 and be done with it. We will need to subtract out the number of students that are in both.

17. Permutations 10 Band Soccer 80 40 In this case, a student could be a member of both groups: If there are 10 students involved in both groups, then we have 40 + 80 – 10, or 110 possibilities for a representative.

18. Permutations Track Soccer 60 40 In the case of soccer and track, the two groups are mutually exclusive, or disjoint, because it is not possible to be part of both groups.

19. Permutations Band Soccer 10 80 40 In the case of soccer and band, the two groups are not mutually exclusive. We must make our adjustment to the addition principle when events are not mutually exclusive.

20. Permutations Say there is a lottery where a player can arrange (order matters) either 2, 3, or 4 numbers from the numbers 1-20. (Naturally the player who matches 4 numbers wins more money than the player who only matches 2 numbers.) In some instances we can use the multiplication principle and the addition principle together.

21. Permutations First, consider the number of ways to arrange 2 numbers from the numbers 1-20: We have 20 choices for the first number, and then 19 for the second number, so we have 20*19, or 380 possibilities. In how many ways can a player fill out a card?

22. Permutations The number of ways to arrange 3 numbers: We have 20 choices for the first number, 19 for the second number, and then 18 for the third number, so we have 20*19*18, or 6,840 possibilities. In how many ways can a player fill out a card?

23. Permutations The number of ways to arrange 4 numbers: 20*19*18*17 = 116,280 possibilities. In how many ways can a player fill out a card?

24. Permutations Finally, to determine the number of ways to fill out a card, we need to add together the possibilities for 2 numbers, for 3 numbers, and for 4 numbers: 380 + 6,840 + 116,280 = 123,500 ways. Can you see why the person matching 4 numbers would win so much more money than someone matching 2 numbers? In how many ways can a player fill out a card?

25. Permutations Permutations are the arrangements of a group of items where order matters. We often calculate permutations in a scenario by using the multiplication principle. We have now seen several examples of what are called permutations.

26. Permutations You have 3 options for your first movie. After choosing the first movie, you will have 2 options remaining for the second movie. After choosing the first and second movies, only 1 option remains for the third movie. This gives you 3*2*1, or 6 ways to arrange 3 movies. Example: In how many ways can choose to watch 3 movies?

27. Permutations The manager has 9 options for the leadoff hitter, then 8 options for the second hitter, then 7 options for the third hitter, and so forth. In total, the manager has 9*8*7*6*5*4*3*2*1, or 362,880 possible lineups. Example: In how many ways can a baseball manager arrange a hitting order for 9 batters?

28. Permutations The teacher has 20 options for the first desk, then 19 options for the second desk, then 18 options for the third desk, and so forth. In total, the teacher has 20*19*18*17*…*5*4*3*2*1, or 2,432,902,008,176,640,000 possibilities. Notice that the answers can get quite large, very quickly. Example: How many seating charts could a teacher make for 20 students, if there are 20 desks in the room?

29. Permutations We usually represent a number this large by using scientific notation: 2,432,902,008,176,640,000 might be rounded to 2.43*1018. Notice that the answers can get quite large, very quickly. Example: How many seating charts could a teacher make for 20 students, if there are 20 desks in the room?

30. Permutations When dealing with extended multiplications as in the previous examples, wouldn’t it be nice to have a shortcut?

31. Permutations Great news—we do! When dealing with extended multiplications as in the previous examples, wouldn’t it be nice to have a shortcut?

32. Permutations Example: The number of ways to arrange 3 movies: 3*2*1 = 3! (read 3 factorial) Do you remember using factorials back in your days of studying Algebra? We can use them again in these problems.

33. Permutations Example: Seating charts for 20 students with 20 desks: 20*19*18*17*…*5*4*3*2*1 = 20! Example: Arranging a batting order for 9 players: 9*8*7*6*5*4*3*2*1 = 9!

34. Permutations One reason this notation is so useful is because most calculators (as well as computer spreadsheets) have a factorial function included, making the calculations quite simple.

35. Permutations On a typical graphing calculator, look for a button labeled “MATH.” After pressing that button, look for a heading labeled “PROBABILITY” or something like “PRB” for short. You might need to use the right arrow to get to that section. Once in the probability section, look for the factorial function (!).

36. Permutations To determine the value of 9! using the calculator follow these steps: Enter 9 Press MATH Move to the probability section Move the cursor to ! Press ENTER—you should see 9! Press ENTER again to get the answer.

37. Permutations The previous examples of permutations (movies, batting orders, and seating charts) share a common feature that makes them relatively simple: they arranged all of the available items. This won’t always be the case.

38. Permutations In this case, we cannot simply say the answer is 4!, because we are choosing from more than 4 movies. Example: You want to pick 4 movies out of 10, and then select an order for viewing them.

39. Permutations We have 10 options for the first movie. After picking that, we have 9 options for the second movie, 8 options for the third movie, and finally 7 options for the fourth movie. Example: You want to pick 4 movies out of 10, and then select an order for viewing them.

40. Permutations This gives us 10*9*8*7, or 5,040 total possibilities. Example: You want to pick 4 movies out of 10, and then select an order for viewing them.

41. Permutations When only ordering 4 items, this is no headache. With a larger number of items, though, it would be nice to have a more compact formula. Example: You want to pick 4 movies out of 10, and then select an order for viewing them.

42. Permutations Notice that we can rewrite 10*9*8*7 as 10*9*8*7*6*5*4*3*2*1 6*5*4*3*2*1 (We multiply by 6*5*4*3*2*1, and then divide by the same amount.) Example: You want to pick 4 movies out of 10, and then select an order for viewing them.

43. Permutations This sneaky little trick allows us to rewrite 10*9*8*7 as 10! 6! Example: You want to pick 4 movies out of 10, and then select an order for viewing them.

44. Permutations Notice the numerator is the factorial of the number of items available, while the denominator is the factorial of the difference between the number of items available and the number of items we’re arranging. 10! 6! Example: You want to pick 4 movies out of 10, and then select an order for viewing them. __10!__ (10-4)! =

45. Permutations In general, if we have n items and want to arrange m of them, we will have __n!__ (n-m)! permutations.

46. Permutations Several other notations are commonly used to express this value: P(n, m) nPm

47. Permutations Graphing calculators often use a function labeled nPr The user enters the total number of items, finds the nPr function, and then enters the number of items to be arranged. (Example to follow.)

48. Permutations Example: A baseball manager wants to arrange a batting lineup for 9 players, and has 15 players from which to choose. How many batting orders can the manager create? One solution would be to multiply the individual options: 15*14*13*12*11*10*9*8*7

49. Permutations Example: A baseball manager wants to arrange a batting lineup for 9 players, and has 15 players from which to choose. How many batting orders can the manager create? Another solution, and one which will prove useful in problems with larger numbers, would be to use our formula for permutations:

50. Permutations Example: A baseball manager wants to arrange a batting lineup for 9 players, and has 15 players from which to choose. How many batting orders can the manager create? P(15, 9) = __15!__ (15-9)! 15! 6! =