AoPS:

1. AoPS: Introduction to Counting & Probability

2. Chapter 5 More with Combinations

3. Intro We’ll see how to do some more complicated problems using combinations and how to use combinations together with some of our other counting methods, such as casework and complementary counting. After that, we’ll move on to the concept of distinguishablilty.

4. Paths on a Grid Problem 5.1: Each block on the grid shown on the next page is 1 unit by 1 unit. Suppose we wish to walk from A to B via a 7 unit path, but we have to stay on the grid – no cutting across blocks. • How many steps to the right do we have to take? • How many steps up do we have to take? • How many different paths can we take?

5. Paths on a Grid We know that we must take a 7 unit path. B A Path consists of 4 steps right and 3 steps up and those steps can be taken in any order. 7! 7 4!3! 3 = = 35

6. Paths on a Grid We know that we must take a 7 unit path. B A So the # of paths is 7C3 = 7 x 6 x 5 = 35. 3 x 2 x 1

7. Exercises 5.2.1 How many paths are there from A to B on the grid shown, if every step must be up or to the right? B A

8. Exercises 5.2.1 How many paths are there from A to B on the grid shown, if every step must be up or to the right? B A There are 5 steps to the right and 2 steps up. These 7 steps can be made in any order, so the answer is 7 7 x 6 2 2 x 1 = = 21

9. Exercise 5.2.2 How many paths are there from C to D on the grid shown, if every step must be down or to the right? C D

10. Exercise 5.2.2 There are 4 steps to the right and 6 steps down. These 10 steps can be made in any order, so 10 10 x 9 x 8 x 7 4 4 x 3 x 2 x 1 C D = = 210

11. Exercise 5.2.3 (a) How many 9-step paths are there from E to G? E G

12. Exercise 5.2.3 • How many 9-step paths are there from E to G? There are 5 steps right and 4 steps down. These 9 steps can be in any order, so 9 9 x 8 x 7 x 6 4 4 x 3 x 2 x 1 E G = = 126

13. Exercise 5.2.3 (b) How many of those paths pass through F? E F G

14. = = 4 different paths From E to F, it is 3 steps to the right and 1 step down, for a total of 4 4 1 1 E F G

15. = = 10 different paths From F to G, it is 2 steps to the right and 3 steps down, for a total of 5 5 x 4 2 2 x 1 E F G

16. = = 10 different paths From F to G, it is 2 steps to the right and 3 steps down, for a total of 5 5 x 4 2 2 x 1 So there are 4 x 10 = 40 paths from E to G through F. E F G

17. More Committee-type Problems Problem 5.2 Coach Grunt is preparing the 5- person starting lineup for his basketball team, the Grunters. There are 12 players on the team. Two of them, Ace & Zeppo, are league All-Stars, so they’ll definitely be in the starting lineup. How many different starting lineups are possible?

18. Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace & Zeppo have been placed in the lineup. The order does not matter, so the answer is 10 10 x 9 x 8 3 3 x 2 x 1 This is a basic combination problem that you should already know! = = 120

19. Problem 5.3 There are 30 men and 40 women in the Town Library Club. They wish to form a 7-person steering committee with 3 men & 4 women. In how many ways can they form the committee?

20. There are 30 men and 40 women in the Town Library Club. They wish to form a 7-person steering committee with 3 men & 4 women. In how many ways can they form the committee? We are selecting 2 separate committees: 3 men from 30 men total in 30 30 x 29 x 28 3 3 x 2 x 1 = = 4,060 ways, and

21. There are 30 men and 40 women in the Town Library Club. They wish to form a 7-person steering committee with 3 men & 4 women. In how many ways can they form the committee? 4 women from 40 women total in 40 40 x 39 x 38 x 37 4 4 x 3 x 2 x 1 = = 91,390 ways

22. = = 91,390 ways 4 women from 40 women total in 40 40 x 39 x 38 x 37 4 4 x 3 x 2 x 1 The 2 committees are independent so we can multiply them together to get the # of ways that we can form the 7-member committee: (4,069) (91,390) = 371,043,400.

23. Problem 5.4 Coach Grunt’s rival team is the Screamers, coached by Coach Yellsalot. The Screamers also have 12 players, but two of them, Bob and Yogi, refuse to play together. How many starting line- ups (of 5 players) can Coach Yellsalot make if the starting lineup can’t contain both Bob and Yogi?

24. Solution 5.4 There are at least 2 ways to solve this: casework or using complementary counting. 3 different cases: Case 1: Bob starts (and Yogi doesn’t) Then the coach must choose 4 more players from the remaining 10 players. So there are 10 lineups that the coach can choose 4

25. Solution 5.4 There are at least 2 ways to solve this: casework or using complementary counting. 3 different cases: Case 2: Yogi starts (and Bob doesn’t) Then the coach must choose 4 more players from the remaining 10 players. So there are 10 lineups that the coach can choose 4

26. Case 3: Neither Bob nor Yogi Starts Then the coach must choose all 5 players from the remaining 10 players. So there are 10 lineups in this case. 5 To get the total # of starting lineups, add each case: 10 10 10 4 4 5 + + = 210 + 210 + 252 = 672

27. Complementary Counting If there are no restrictions then the coach must choose all 5 players from the entire roster of 12 players which he can do in 12 ways. 5 But then we have to subtract the lineups that are not allowed, which are the lineups in which both Bob & Yogi start. This was counted in problem 5.2: the coach must choose 3 more players from the remaining 10 players to complete the lineup, and he can in 10C3 ways.

28. - = 792 – 120 = 672 So the answer to the problem is the total # of lineups (without restrictions) minus the # of lineups that are not allowed: 12 10 5 3

29. Problem 5.5 In how many ways can a dog breeder separate his 10 puppies into a group of 4 and a group of 6 if he has to keep Biter and Nipper, two of the puppies, in separate groups?

30. Since Biter and Nipper cannot be in the same group, subtract the # of ways to form 2 groups with Biter and Nipper in the same group. This means Casework. Case 1: Biter & Nipper in the same group If they are both in the smaller group, we have to choose 2 more dogs from the 8 remaining to complete the smaller group, 8C2 ways or 8 . 2

31. Warning! Don’t mistakenly count the possibilities in Case 1 as 8 8 by reasoning that we must choose 2 of 8 for 26 the smaller group, and 6 from 8 for the larger group. These choices are not independent! Once we pick the 2 dogs for the smaller group, there is no choice but to put the remaining 6 dogs into the larger group.

32. Case 2: Biter & Nipper are in the larger group If both are in larger group, then choose 4 dogs from the remaining 8 to compose the smaller group which can be done 8C4 ways. To get the # of ways to form groups such that Biter & Nipper are both in the same group, add the counts from our 2 cases, 8 + 8 24

33. Case 2: Biter & Nipper are in the larger group Since these are the cases we don’t want, subtract this count from the # of ways to form the 2 groups w/o restrictions. 10 - 8 + 8 4 24 =210 – (28 + 70) =112

34. The other way is to solve this by direct casework. Case 1: Biter is in smaller group, Nipper is in larger. To complete smaller, choose 3 more dogs from the remaining 8, 8C3. Case 2: Nipper is in larger group, Bitter is in smaller. Again to complete the smaller, choose 3 more dogs from the remaining 8, 8 3

35. So to get the total count, add 56 + 56 = 112.

36. Exercises 5.3.1 A Senate committee has 8 Republicans & 6 Democrats. In how many ways can we form a sub- committee with 3 Republicans & 2 Democrats?

37. Solution 5.3.1 A Senate committee has 8 Republicans & 6 Democrats. In how many ways can we form a sub- committee with 3 Republicans & 2 Democrats? There are 8 Rep’s and 3 spots for them, so there are 8 = 56 ways to choose Rep’s. 3 There are 6 Dem’s and 2 spot for them, so there are 6 = 15 ways to choose Dem’s. 56 x 15 = 840 ways 2

38. Exercises • 5.3.2 Our school’s girls volleyball team has 14 • players, including a set of 3 triplets: Alicia, Amanda, • & Anna. In how many ways can we choose 6 starters • with no restrictions? • if all 3 triplets are in the starting lineup? • if exactly one of the triplets is in the starting • lineup? • (d) if at most one of the triplets is in the starting • lineup?

39. Solution • 5.3.2 Our school’s girls volleyball team has 14 • players, including a set of 3 triplets: Alicia, Amanda, • & Anna. In how many ways can we choose 6 starters • with no restrictions? • Choosing 6 starters from 14 players, • 14 = 3003 ways. • 6

40. 5.3.2 Our school’s girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, & Anna. In how many ways can we choose 6 starters (b) if all 3 triplets are in the starting lineup? If all triplets are in the starting lineup, choose the 3 remaining starters from the 11 players, 11 = 165 ways. 3

41. 5.3.2 Our school’s girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, & Anna. In how many ways can we choose 6 starters (c) if exactly one of the triplets is in the starting lineup? If exactly 1 is in the lineup, we have 3 choices for which triplet to put in the starting lineup, and then 11 people to choose for the remaining spots. So 3 x 11 = 3 x 462 = 1386. 5

42. In how many ways can we choose 6 starters (d) if at most one of the triplets is in the starting lineup? Add together the # of lineups with one triplet & with no triplets. The # of lineups with no triplets is 11 6 = 462, since we must choose 6 starters from the 11 remaining players. When one triplet is in the lineup, there are 1386 options (part c). So the total is 1386 + 462 = 1848.

43. Exercises 5.3.3 Suppose we want to divide the 10 dogs from Problem 5.5 into 3 groups, one with 3 dogs, one with 5 dogs, and one with 2 dogs. How many ways can we form the groups such that Biter is in the 3-dog group and Nipper is in the 5-dog group?

44. Solution 5.3.3 Suppose we want to divide the 10 dogs from Problem 5.5 into 3 groups, one with 3 dogs, one with 5 dogs, and one with 2 dogs. How many ways can we form the groups such that Biter is in the 3-dog group and Nipper is in the 5-dog group? Place Biter in the 3-dog group & Nipper in the 5-dog group. This leaves 8 dogs remaining to put in the last 2 spots of Biter’s group, which is 8 ways. 2

45. Solution Place Biter in the 3-dog group & Nipper in the 5-dog group. This leaves 8 dogs remaining to put in the last 2 spots of Biter’s group, which is 8 ways. 2 Then there are 6 dogs remaining for the last 4 spots in Nipper’s group, which is 6 ways. 2

46. Solution The remaining 2-dog group takes the last 2 dogs. So the total # of possibilities is 8 x 6 = 420. 2 4

47. Exercises 5.3.4 We call a number a descending number if each digit is strictly smaller than the digit that comes before it. For example, 863 is a descending #. How many 3-digit descending numbers are there?

48. Solution 5.3.4 We call a number a descending number if each digit is strictly smaller than the digit that comes before it. For example, 863 is a descending #. How many 3-digit descending numbers are there? For every 3 different digits, there is one corresponding descending #, which is just the digits in descending order. So the answer is the # of combinations of 3 different digits, which is 10 = 120. 3

49. Exercises 5.3.5 We call a number a mountain number if its middle digit is strictly larger than any other digit. before it. For example, 284 is a mountain #. How many 3-digit mountain numbers are there?

50. Solution 5.3.5 We call a number a mountain number if its middle digit is strictly larger than any other digit. before it. For example, 284 is a mountain #. How many 3-digit mountain numbers are there? Case 1: numbers of the form xyz (x ≠ 0) Any pair of nonzero digits has a corresponding palindrome (xyz) mountain #, so the # of these is 9 = 36. 2