1 / 87

890 likes | 1.22k Views

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-

Download Presentation
## AoPS:

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**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

More Related