Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
AoPS: Introduction to Counting & Probability Chapter 5 More with Combinations 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. 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. (a)How many steps to the right do we have to take? (b)How many steps up do we have to take? (c)How many different paths can we take? Paths on a Grid We know that we must take a 7 unit path. B A 7! = 7 = 35 4!3! 3 Path consists of 4 steps right and 3 steps up and those steps can be taken in any order. 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. 3x2x1 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 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 7x6 = = 21 2 2x1 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 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 C 10 = 10 x 9 x 8 x 7 = 210 4 4x3x2x1 D Exercise 5.2.3 (a) How many 9-step paths are there from E to G? E G Exercise 5.2.3 (a) How many 9-step paths are there from E to G? E 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 = 126 4 4x3x2x1 Exercise 5.2.3 (b) How many of those paths pass through F? E F G From E to F, it is 3 steps to the right and 1 step down, for a total of 4 4 = = 4 different paths 1 1 E F G From F to G, it is 2 steps to the right and 3 steps down, for a total of 5 5 x 4 = = 10 different paths 2 2x1 E F G From F to G, it is 2 steps to the right and 3 steps down, for a total of 5 5 x 4 = = 10 different paths 2 2x1 E F So there are 4 x 10 = 40 paths from E to G through F. G More Committee-type Problems Problem 5.2 Coach Grunt is preparing the 5person 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? 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 = 120 3 3x2x1 This is a basic combination problem that you should already know! 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? 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 = 4,060 ways, and 3 3x2x1 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 = 91,390 ways 4 4x3x2x1 4 women from 40 women total in 40 = 40 x 39 x 38 x 37 = 91,390 ways 4 4x3x2x1 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. 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 lineups (of 5 players) can Coach Yellsalot make if the starting lineup can’t contain both Bob and Yogi? 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 4 that the coach can choose 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 4 that the coach can choose 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 5 in this case. To get the total # of starting lineups, add each case: 10 4 10 + 4 10 + 5 = 210 + 210 + 252 = 672 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. So the answer to the problem is the total # of lineups (without restrictions) minus the # of lineups that are not allowed: 12 5 10 - 3 = 792 – 120 = 672 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? 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 Warning! Don’t mistakenly count the possibilities in Case 1 as 8 8 by reasoning that we must choose 2 of 8 for 2 6 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. 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 2 4 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 4 - 8 2 + 8 4 =210 – (28 + 70) =112 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 So to get the total count, add 56 + 56 = 112. Exercises 5.3.1 A Senate committee has 8 Republicans & 6 Democrats. In how many ways can we form a subcommittee with 3 Republicans & 2 Democrats? Solution 5.3.1 A Senate committee has 8 Republicans & 6 Democrats. In how many ways can we form a subcommittee 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 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 (a)with no restrictions? (b) if all 3 triplets are in the starting lineup? (c) if exactly one of the triplets is in the starting lineup? (d) if at most one of the triplets is in the starting lineup? 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 (a)with no restrictions? Choosing 6 starters from 14 players, 14 = 3003 ways. 6 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 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 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. 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? 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 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 Solution The remaining 2-dog group takes the last 2 dogs. So the total # of possibilities is 8 x 6 = 420. 2 4 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? 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 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? 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 2 = 36. 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 2: numbers of the form xyz (z ≠ 0, x ≠ z) Any group of 3 nonzero digits (y > x > z > 0) has 2 corresponding mountain #’s (xyz and zyx), so the # of these is 2x 9 3 = 168. 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 3: numbers of the form xy0 (x ≠ 0, y ≠ 0) Any pair of nonzero digits has a corresponding mountain # in the form xy0, so there are 9 2 = 36. 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? So the total # of mountain numbers is 36 + 168 + 36 = 240. Distinguishability Distinguishable means to tell items apart and indistinguishable means items cannot be told apart. A B C D Box 1 4 distinguishable balls and Box 2 2 distinguishable boxes Distinguishability Distinguishable means to tell items apart and indistinguishable means items cannot be told apart. A B C D 4 distinguishable balls and 2 indistinguishable boxes Distinguishability Distinguishable means to tell items apart and indistinguishable means items cannot be told apart. Box 1 4 indistinguishable balls and Box 2 2 distinguishable boxes Distinguishability Distinguishable means to tell items apart and indistinguishable means items cannot be told apart. 4 indistinguishable balls and 2 indistinguishable boxes Problem 5.6 How many ways are there to put 4 distinguishable balls into 2 distinguishable boxes? Solution For each ball, there are 2 choices of which box to place it in. Since this choice is independent for each of the 4 balls, multiply the # of choices together. A B C D Box 1 Box 2 So there are 24 = 16 ways to place 4 distinguishable balls into 2 distinguishable boxes. Problem 5.7 How many ways are there to put 4 distinguishable balls into 2 indistinguishable boxes? Problem 5.7 How many ways are there to put 4 distinguishable balls into 2 indistinguishable boxes? A B C D Problem 5.7 We don’t care which box is which, we only care about which balls are together and which ones aren’t. There are 2 ways we can do this count. 1st: take the count from Prob 5.6 & divide by the # of ways to arrange the boxes – 2! = 2 ways so A B C D there are 16/2 = 8 ways to arrange 4 distinguishable balls into 2 indistinguishable boxes. Problem 5.7 WARNING: This method does not generalize to more than 2 boxes, as you will see in Exercise 5.4.2. A B C D Problem 5.7 Alternatively, this could have been solved by casework. Case 1: One box has 4 balls, the other has 0 balls There is only 1 way to do this – put all the balls into A B C D one box (it doesn’t matter which box, since they are indistinguishable). Problem 5.7 Alternatively, this could have been solved by casework. Case 2: One box has 3 balls, the other has 1 ball The choice here is which ball is by itself in a box & A B C D there are 4 choices. Problem 5.7 Case 3: Each box has 2 balls At 1st glance it looks like we have to choose 2 balls from the total of 4 balls to go into the 1st box, & then the remaining 2 balls will go into the 2nd box. A B C D So there are 4C2 = 6 choices. However, this overcounts the choices by a factor of 2. Problem 5.7 Case 3: Each box has 2 balls For Example: Look at balls A, B, C, & D. If we 1st choose A & B, then one box has A, B, and the other has C, D. A B C D Problem 5.7 Case 3: Each box has 2 balls But this is exactly the same situation as if we had originally put C, D in the 1st box and A, B in the 2nd box, A B C D so divide by 2, giving 6/2 = 3 possibilities Problem 5.7 To get the total number of possibilities, add the counts from the 3 cases: 1 + 4 + 3 = 8 ways to arrange 4 distinguishable balls into 2 indistinguishable boxes. A B C D Problem 5.7 WARNING: This casework approach is not quite as simple if there are 3 or more boxes, as will be demonstrated in Exercise 5.4.2. A B C D Problem 5.8 How many ways are there to put 4 indistinguishable balls into 2 distinguishable boxes? Box 1 Box 2 Problem 5.8 How many ways are there to put 4 indistinguishable balls into 2 distinguishable boxes? Since the balls are indistinguishable, the only thing to keep track of is how many balls are in each box. In this case, the cases can be listed: Box 1 Box 2 Problem 5.8 How many ways are there to put 4 indistinguishable balls into 2 distinguishable boxes? Since the balls are indistinguishable, the only thing to keep track of is how many balls are in each box. In this case, the cases can be listed: Box 1 Box 2 put either 0, 1, 2, 3, 4 balls into the 1st box, & the rest into the 2nd box. Problem 5.8 Since the balls are indistinguishable, the only thing to keep track of is how many balls are in each box. In this case, the cases can be listed: put either 0, 1, 2, 3, 4 balls into the 1st box, & the rest into the 2nd box. Box 1 Box 2 So there are 5 ways to arrange 4 indistinguishable balls into 2 distinguishable boxes. Problem 5.9 How many ways are there to put 4 indistinguishable balls into 2 indistinguishable boxes? In this problem, it is only necessary to count the # of ways to split 4 items into 2 groups. There are only 3 ways: {4, 0}, {3, 1} & {2, 2} So there are only 3 ways to put 4 indistinguishable balls into 2 indistinguishable boxes. In these four problems, it is clear that distinguishability matters. Nothing to memorize. You should be able to use logic on any given problem to figure out how to account for the distinguishability and/or indistinguishability of what you are counting. The purpose of problems 5.6-5.9 is to make it clear that there is a difference. Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (a) the balls are distinguishable and the boxes are distinguishable? (b) the balls are distinguishable but the boxes are not? (c) the balls are not distinguishable but the boxes are (d) the balls are not distinguishable and neither are the boxes? Solutions 5.4.1 How many ways are there to put 5 balls in 2 boxes if (a) the balls are distinguishable and the boxes are distinguishable? There are 2 different boxes, so each of the 5 balls can be placed in 2 different locations. So the answer is 25 = 32. Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (b) the balls are distinguishable but th boxes are not? Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the # of balls in each box: Case 1: 5 balls in one box, 0 in the other box. We must choose 5 balls to go in one box, which can be done in 5C5 ways = 1 way. Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (b) the balls are distinguishable but th boxes are not? Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the # of balls in each box: Case 2: 4 balls in one box, 1 in the other box. We must choose 4 balls to go in one box, which can be done in 5C4 ways = 5 ways. Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (b) the balls are distinguishable but th boxes are not? Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the # of balls in each box: Case 3: 3 balls in one box, 2 in the other box. We must choose 3 balls to go in one box, which can be done in 5C3 ways = 10 ways. Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (b) the balls are distinguishable but th boxes are not? This gives a total of 1 + 5 + 10 = 16 arrangements. Also note that every arrangement of balls when the boxes are indistinguishable is counted twice in the distinguishable case. So simply divide the answer from part (a) by 2. However, this doesn’t work if there’s more than 2 boxes (as we’ll see later). Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (c) the balls are distinguishable but th boxes are not? Since the balls are indistinguishable, we need only count the # of balls in the distinguishable boxes. we can put 5, 4, 3, 2, 1, or 0 balls in Box #1 (and the rest go in Box #2). So there are 6 different arrangements. Exercises 5.4.1 How many ways are there to put 5 balls in 2 boxes if (d) the balls are distinguishable but th boxes are not? Since both balls & boxes are indistinguishable, we can arrange them with 5 in one and 0 in the other, 4 in one and 1 in the other, or 3 in one and 2 in the other, for a total of 3 different arrangements. Exercise 5.4.2 How many ways are there to put 5 balls in 3 boxes (a) the balls are distinguishable and the boxes are distinguishable? (b) the balls are distinguishable but the boxes aren’t (c) the balls are not distinguishable but the boxes are (d) the balls are not distinguishable and neither are the boxes Exercise 5.4.2 How many ways are there to put 5 balls in 3 boxes (a) the balls are distinguishable and the boxes are distinguishable? There are 3 different boxes, so each of the 5 balls can be placed in 3 different locations, so the answer is 35 = 243. Exercise 5.4.2 How many ways are there to put 5 balls in 3 boxes (b) the balls are distinguishable but the boxes aren’t Since the boxes are indistinguishable, there are 5 different cases for arrangements of the # of balls in each box: (5,0,0), (4,1,0), (3,2,0), (3,1,1), or (2,1,1) (b) the balls are distinguishable but the boxes aren’t Since the boxes are indistinguishable, there are 5 different cases for arrangements of the # of balls in each box: (5,0,0), (4,1,0), (3,2,0), (3,1,1), or (2,1,1) (5,0,0): There is only 1 way to put all 5 balls in one box (b) the balls are distinguishable but the boxes aren’t Since the boxes are indistinguishable, there are 5 different cases for arrangements of the # of balls in each box: (5,0,0), (4,1,0), (3,2,0), (3,1,1), or (2,1,1) (5,0,0): There is only 1 way to put all 5 balls in one box (4,1,0): There are 5C4 = 5 choices for 4 balls in one box (b) the balls are distinguishable but the boxes aren’t Since the boxes are indistinguishable, there are 5 different cases for arrangements of the # of balls in each box: (5,0,0), (4,1,0), (3,2,0), (3,1,1), or (2,1,1) (5,0,0): There is only 1 way to put all 5 balls in one box (4,1,0): There are 5C4 = 5 choices for 4 balls in one of the boxes (3,2,0): There are 5C3 = 10 choices for 3 balls in one of the boxes (b) the balls are distinguishable but the boxes aren’t (5,0,0), (4,1,0), (3,2,0), (3,1,1), or (2,1,1) (5,0,0): There is only 1 way to put all 5 balls in one box (4,1,0): There are 5C4 = 5 choices for 4 balls in one of the boxes (3,2,0): There are 5C3 = 10 choices for 3 balls in one of the boxes (3,1,1): There are 5C3 = 10 choices for 3 balls in one of the boxes (2,2,1): There are 5C2 = 10 options for one of the boxes with 2 balls, then 3C2 = 3 options for the 2nd box with 2 balls, & one option remaining for the 3rd. But, since the boxes with 2 balls are indistinguishable, we are counting each pair of balls twice, and must divide by 2. So there are (10 X 3) ÷ 2 = 15 arrangements of balls as (2,2,1). Thus the total # of arrangements for 3 indistinguishable boxes and 5 distinguishable balls is 1 + 5 + 10 + 10 + 15 = 41. Exercise 5.4.2 How many ways are there to put 5 balls in 3 boxes (c) the balls are not distinguishable but the boxes are Since the balls are indistinguishable, we must only count the # of balls in the different boxes. There are 3 ways to arrange the balls as (5, 0, 0): box 1 can have 5, box 2 can have 5, box 3 can have 5. There are 3! = 6 ways to arrange (4, 1, 0) and 3! = 6 ways to arrange (3, 2, 0); in each case we must choose 1 of the 3 boxes to have the largest # of balls, and also one of the remaining 2 boxes to be left empty. However, there are only 3 ways to arrange (3, 1, 1) and 3 ways to arrange (2, 2, 1); in each case, we must choose one box to have the “different” # of balls (3 in the (3, 1, 1) case and 1 in the (2, 2, 1) case). This gives a total of 3 + 6 + 6 + 3 + 3 = 21 arrangements. Exercise 5.4.2 How many ways are there to put 5 balls in 3 boxes (d) the balls are not distinguishable and neither are the boxes The ways to arrange indistinguishable balls into indistinguishable boxes only depends on the # of balls in the boxes. The ways to do this are (5, 0, 0), (4, 1, 0), (3, 2, 0), (3, 1, 1), (2, 2, 1). There are 5 ways. Review Problems 5.10 In the diagram shown (a) how many paths are there from A to B? (Assume all paths only go up and to the right.) (b) how many paths are there from A to C? (c) how many paths are there from C to B? (d) how many paths are there from A to B passing through C? B C A Solutions 5.10 In the diagram shown (a)how many paths are there from A to B? (Assume all paths only go up and to the right.) There are 5 steps to the right and 4 steps up. These 9 steps can be made in an order, so the answer is 9C4 = 126. B C A Solutions 5.10 In the diagram shown (b) how many paths are there from A to C? There is 1 step to the right and 2 steps up. These 3 steps can be made in any order, so the answer is 3C1 = 3. B C A Solutions 5.10 In the diagram shown (c) how many paths are there from C to B? There are 4 steps to the right and 2 steps up. These 6 steps can be made in any order, so the answer is 6C4 = 15. B C A Solutions 5.10 In the diagram shown (d) how many paths are there from A to B passing through C? There are 3 paths from A to C and 15 paths from C to B, so there are 3 X 15 = 45 paths from A to B through C. B C A Review Problems 5.13 My school’s math club has 6 boys and 8 girls. I need to select a team to send to the state math competition. We want 6 people on the team. In how many ways can I select the team: (a)without restrictions? (b)to have 3 boys and 3 girls? (c)to have more girls than boys? Solutions We want 6 people on the team. In how many ways can I select the team: (a)without restrictions? With no restrictions, we are merely picking 6 out of 14. This is 14C6 = 3003. Solutions We want 6 people on the team. In how many ways can I select the team: (b) to have 3 boys and 3 girls? We are picking 3 boys out of 6, so there are 6C3 = 20 options for the boys on the team. We are picking 3 girls out of 8, so there are 8C3 = 56 options for the girls on the team. This gives a total of 20 X 56 = 1120 choices. Solutions (c) to have more girls than boys? We do this problem similarly to part (b), except with 3 cases. Case 1: 4 girls, 2 boys on the team. With 4 girls on the team, there are 8C4 = 70 ways to pick the girls, & 6C2 = 15 ways to pick the boys for a total of 70 X 15 = 1050. Solutions (c) to have more girls than boys? We do this problem similarly to part (b), except with 3 cases. Case 2: 5 girls, 1 boy on the team. With 5 girls on the team, there are 8C5 = 56 ways to pick the girls, & 6C1 = 6 ways to pick the boy for a total of 56 X 6 = 336. Solutions (c) to have more girls than boys? We do this problem similarly to part (b), except with 3 cases. Case 3: 6 girls on the team. With 6 girls on the team, there are 8C6 = 28 ways to pick the girls on the team. This gives us a sum of 1050 + 336 + 28 = 1414. Review Problems 5.14 The Sagebrush student council has 6 boys and 6 girls as class representatives. Two committees, each consisting of 2 boys & 2 girls, are to be created. If no student can serve on both committees, how many different combinations of committees are possible? (Source: MATHCOUNTS) Solution To make the 1st committee, we choose 2 boys in 6C2 = 15 ways and 2 girls in 6C2 = 15 ways, for a total of 15 x 15 = 225 possible committees. Then to make the 2nd committee, we choose 2 more boys in 4C2 = 6 ways & 2 more girls in 4C2 = 6 ways for a total of 6 x 6 possibilities. So the # of ways to form 2 committees like this is 225 x 36 = 8100. Solution But this counts each pair of committees twice (in other words, there is no “first” committee or “second” committee), so we must divide by 2 to get the final answer of 8100 ÷ 2 = 4050 pairs of committees. Challenge Problem 5.22 Twenty married couples are at a party. Every man shakes hands with everyone except himself and his spouse. Half of the women refuse to shake hands with any other women. The other 10 women all shake hands with each other (but not with themselves). How many handshakes are there at the party? Solution 20 men shake 38 hands each (not themselves or spouses), so combined they shake 760 hands. 10 women shake only 19 hands (all the men but their husbands and the 9 other friendly women), so women shake a total of 10 x 19 + 10 x 28 = 470 hands. This means a total of 1230 hands have been shaken. However, each handshake has been counted twice, so we divide by 2 and see that there are 615 handshakes. Challenge 5.23 Five different circles are drawn in a plane. What is the maximum number of different points at which the circles can meet? Solution Each circle can meet each other circle in at most 2 points. Draw five circles such that each circle intersects with every other circle twice, and so that no 3 circles intersect at the same point. Since there are 5C2 = 10 pairs of circles, and 2 intersections per pair of circles, there are 10 x 2 = 20 intersections. This ends Chapter 5