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AoPS:. Introduction to Probability and Counting. Chapter 4. Committees and Combinations. Committee Forming. Problem 4.1 In how many ways can a President and a Vice President be chosen from a group of 4 people (assuming that the President and the Vice-Presi-

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**AoPS:**Introduction to Probability and Counting**Chapter 4**Committees and Combinations**Committee Forming**Problem 4.1 • In how many ways can a President and a Vice President be chosen from a group of 4 people (assuming that the President and the Vice-Presi- dent cannot be the same person)?**Committee Forming**Problem 4.1 This is a permutation problem like those in Chp1. There are 4 choices for Pres. and 3 choices for VP, so there are 4 x 3 = 12 choices.**Committee Forming**Problem 4.1 (b) In how many ways can a 2-person committee be chosen from a group of 4 people (where the order in which we choose the 2 people doesn’t matter)?**Committee Forming**Problem 4.1 There are 4 ways to choose person A & 3 ways to choose person B, but then we’ve overcounted again since choosing person A and then person B will give the same committee as choosing person B and then person A. Each committee is counted twice in the original 4 x 3 count, so divide by 2 to correct for overcount: (4 x 3) /2 = 6 ways to choose the 2 person committee.**Committee Forming**Problem 4.1 Not convinced? Label the people, A, B, C, & D. Then list all the possible ways to select a President & Vice-President.**Committee Forming**Problem 4.1 Pres. VP Committee Pres. VP Committee A B AB B C BC B A C B A C AC B D BD C A D B A D AD C D CD D A D C**Important: Understand the difference between forming**committees (order doesn’t matter) & picking distinct officers (order of selection matters). Pres. VP Committee Pres. VP Committee A B AB B C BC B A C B A C AC B D BD C A D B A D AD C D CD D A D C**Problem 4.2**In how many ways can 3 people be chosen from a group of 8 people to form a committee?**Problem 4.2**There are 8 ways to choose the 1st, 7 ways to choose the 2nd, and 6 ways to choose the 3rd. But since we are choosing a committee, not officers, the order does not matter. This is the key to any committee-forming problems.**Problem 4.2**Suppose the committee is made up of A, B, & C. It can be chosen in any of the following orders: ABC, ACB, BAC, BCA, CAB, CBA so each possible committee corresponds to 3! = 6 possible orderings of officers. Since each committee is overcounted by 3!, the # of committees that can be chosen from 8 people is (8 x 7 x 6) / 3! = 56.**Combinations**Special notations are used for choosing committees. Denote the # of ways we can choose an r-person committee from a total of n people as C(n, r) or n C ror n r**Combinations**Special notations are used for choosing committees. Denote the # of ways we can choose an r-person committee from a total of n people as C(n, r) or n C ror n r This is read as “n choose r” for the # of ways to choose an r-person committee from a total of n people.**Combinations**Looking back at Problems 4.1 and 4.2 4 8 2 3 = 6 = 56**Problem 4.3**In my state’s lottery, 48 balls are numbered 1-48, and 6 are chosen. How many different sets of winnings numbers are there? (In this lottery, the order in which the numbers are chosen does not matter.)**Problem 4.3**This is the same as forming a committee – instead of people, we have balls. 48 x 47 x 46 x 45 x 44 x 43 6! = 12,271,512**Problem 4.3**This is the same as forming a committee – instead of people, we have balls. 48 x 47 x 46 x 45 x 44 x 43 6! This number is also 48 6 This means your chance of winning the lottery is 1 in 12 million. = 12,271,512**Exercises**4.2.1 (a) In how many ways can I choose 4 different officers from a club of 9 people? (b) In how many ways can I choose a 4- person committee from a club of 9 people?**Exercises**4.2.1 (a) In how many ways can I choose 4 different officers from a club of 9 people? 3024 (b) In how many ways can I choose a 4- person committee from a club of 9 people? 126**Exercises**4.2.2 My club has 25 members. In how many ways can I choose members to form a 4- person executive committee?**Exercises**4.2.2 My club has 25 members. In how many ways can I choose members to form a 4- person executive committee? 12, 650**Exercises**4.2.3 Our water polo team has 15 members. I want to choose a starting line-up consisting of 7 players, one of whom will be the goalie (the other six positions are interchangeable). In how many ways can I choose my starting line-up?**Exercises**4.2.3 Our water polo team has 15 members. I want to choose a starting line-up consisting of 7 players, one of whom will be the goalie (the other six positions are interchangeable). In how many ways can I choose my starting line-up? 45,045**Exercises**4.2.4 Consider a regular octagon. How many triangles can be formed whose vertices are the vertices of the octagon?**Exercises**4.2.4 Consider a regular octagon. How many triangles can be formed whose vertices are the vertices of the octagon? No 3 vertices are collinear so any combination of 3 vertices will from a triangle, so there are 8 x 7 x 6 3! = 56**Exercises**4.2.5 The Senate has 100 members, consisting of 55 Republicans and 45 Democrats. In how many ways can I choose a 5-person committee consisting of 3 Republicans and 2 Democrats?**Exercises**4.2.5 The Senate has 100 members, consisting of 55 Republicans and 45 Democrats. In how many ways can I choose a 5-person committee consisting of 3 Republicans and 2 Democrats? There are 55 x 54 x 53 ways to choose R’s 3! and 45 x 44 ways to choose D’s. 2! So there are 26,235 x 990 = 25,972,650 ways to choose a committee. = 26,235 = 990**Exercises**4.2.6 My state’s lottery has 30 white balls numbered 1 – 30, and 20 red balls numbered 1-20. In each lottery drawing, 3 of the white balls and 2 of the red balls are drawn. To win you must match all 3 white balls and both red balls, without regard to the order in which they were drawn. How many possible different combinations may be drawn?**Exercises**4.2.6 My state’s lottery has 30 white balls numbered 1 – 30, and 20 red balls numbered 1-20. In each lottery drawing, 3 of the white balls and 2 of the red balls are drawn. To win you must match all 3 white balls and both red balls, without regard to the order in which they were drawn. How many possible different combinations may be drawn? Ways to choose white balls: 30 x 29 x 28 3! Ways to choose red balls: 20 x 19 2! = 4060 = 190**Exercises**4.2.6 My state’s lottery has 30 white balls numbered 1 – 30, and 20 red balls numbered 1-20. In each lottery drawing, 3 of the white balls and 2 of the red balls are drawn. To win you must match all 3 white balls and both red balls, without regard to the order in which they were drawn. How many possible different combinations may be drawn? So total # of outcomes for both red and white balls is 4060 x 190 = 771,400.**How To Compute Combinations**Problem 4.4: Consider a club which has n people. • Determine a formula for the number of ways to choose r different officers from n people. (b) Determine the number of ways that any given r people can be assigned to be officers. (c) Using your answers to parts (a) & (b), determine a formula for the number of ways to form an r-person committee from a total of n people.**How To Compute Combinations**Problem 4.4: Consider a club which has n people. Start by counting the # of ways to choose r people if order matters. There are n choices for the 1st person, n – 1 choices for the 2nd person, n – 2 choices for the 3rd, and so on, up to n – r + 1 choices (See p.20, Prob 1.15 of Intro to Counting & Probability.)**How To Compute Combinations**Problem 4.4: Consider a club which has n people. So there are nx (n – 1) x (n – 2) x … x (n – r + 1) ways to choose r people from a total of n people if order matters. This is the quantity that we denote by P(n, r).**How To Compute Combinations**Problem 4.4: Consider a club which has n people. But we know that there are r! ways to order r people. Therefore, each unordered committee of r people will correspond to r! ordered choices of r people. So we need to divide our count by r! to correct for overcounting.**How To Compute Combinations**nn x (n – 1) x (n – 2) x … x (n – r + 1) r r! n! r! (n – r)! = =**Problem 4.5**Compute 11 4**Problem 4.5**= = Compute 11 11! 11 x 10 x 9 x 8 4 4!7! 4 x 3 x 2 x 1 330 SHORTCUT: Put the 1str terms of n! in the numerator and r! in the denominator. =**Problem 4.6**A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during a n-person round-robin tennis tournament, where n > 2?**Problem 4.6**A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during a n-person round-robin tennis tournament, where n > 2? Use combinations. Think of each match in the tournament as a 2-person committee, where the 2 players in the match are a “committee.”**Problem 4.6**Use combinations. Think of each match in the tournament as a 2-person committee, where the 2 players in the match are a “committee.” So the # of matches in the tournament equals the # of ways to choose a 2-person committee out of n people, which is n 2**Exercises**4.3.1 Compute the following combinations: 5 8 10 9 9 3 2 4 8 1**Exercises**4.3.1 Compute the following combinations: 5 10 8 28 10 210 9 9 9 9 3 2 4 8 1**Problem 4.3.2**What is nfor any positive integer n? 0**Problem 4.3.2**What is nfor any positive integer n? 0 0! = 1 by definition, so the combination equals n! 0! n! = 1**Problem 4.3.2**What is nfor any positive integer n? 1**Problem 4.3.2**What is nfor any positive integer n? 1 n! 1!(n -1)! = n**Problem 4.3.2**What is nfor any positive integer n? n**Problem 4.3.2**What is nfor any positive integer n? n n! n! n! n!(n -n)! n!0! n! = 1 = =**Our First Combinatorial Identity**Problem 4.7: Compute • 6 6 2 4 • 88 3 5 and and

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