Permutations, Combinations & Probability: Finding Object Combinations, Assignments of Mathematics

Download Permutations, Combinations & Probability: Finding Object Combinations and more Assignments Mathematics in PDF only on Docsity! 35 Permutations, Combinations and Proba- bility Thus far we have been able to list the elements of a sample space by drawing a tree diagram. For large sample spaces tree diagrams become very complex to construct. In this section we discuss counting techniques for finding the number of elements of a sample space or an event without having to list them. Permutations Consider the following problem: In how many ways can 8 horses finish in a race (assuming there are no ties)? We can look at this problem as a decision consisting of 8 steps. The first step is the possibility of a horse to finish first in the race, the second step the horse finishes second, ... , the 8th step the horse finishes 8th in the race. Thus, by the Fundamental Principle of counting there are 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40, 320 ways This problem exhibits an example of an ordered arrangement, that is, the order the objects are arranged is important. Such ordered arrangement is called a permutation. Products such as 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 can be written in a shorthand notation called factoriel. That is, 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 8! (read ”8 factoriel”). In general, we define n factoriel by n! = { n(n− 1)(n− 2) · · · 3 · 2 · 1, if n ≥ 1 1, if n = 0. where n is a whole number n. Example 35.1 Evaluate the following expressions: (a) 6! (b) 10! 7! . Solution. (a) 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 (b) 10! 7! = 10·9·8·7·6·5·4·3·2·1 7·6·5·4·3·2·1 = 10 · 9 · 8 = 720 Using factoriels we see that the number of permutations of n objects is n!. 1 Example 35.2 There are 6! permutations of the 6 letters of the word ”square.” In how many of them is r the second letter? Solution. Let r be the second letter. Then there are 5 ways to fill the first spot, 4 ways to fill the third, 3 to fill the fourth, and so on. There are 5! such permutations. Example 35.3 Five different books are on a shelf. In how many different ways could you arrange them? Solution. The five books can be arranged in 5 · 4 · 3 · 2 · 1 = 5! = 120 ways. Counting Permutations We next consider the permutations of a set of objects taken from a larger set. Suppose we have n items. How many ordered arrangements of r items can we form from these n items? The number of permutations is denoted by P (n, r). The n refers to the number of different items and the r refers to the number of them appearing in each arrangement. This is equivalent to finding how many different ordered arrangements of people we can get on r chairs if we have n people to choose from. We proceed as follows. The first chair can be filled by any of the n people; the second by any of the remaining (n−1) people and so on. The rth chair can be filled by (n− r+1) people. Hence we easily see that P (n, r) = n(n− 1)(n− 2)...(n− r + 1) = n! (n− r)! . Example 35.4 How many ways can gold, silver, and bronze medals be awarded for a race run by 8 people? Solution. Using the permuation formula we find P (8, 3) = 8! (8−3)! = 336 ways. Example 35.5 How many five-digit zip codes can be made where all digits are unique? The possible digits are the numbers 0 through 9. 2 choosing Ms B and Mr A. A combination is a group of items in which the order does not make a difference. Counting Combinations Let C(n, r) denote the number of ways in which r objects can be selected from a set of n distinct objects. Since the number of groups of r elements out of n elements is C(n, r) and each group can be arranged in r! ways then P (n, r) = r!C(n, r). It follows that C(n, r) = P (n, r) r! = n! r!(n− r)! . Example 35.6 How many ways can two slices of pizza be chosen from a plate containing one slice each of pepperoni, sausage, mushroom, and cheese pizza. Solution. In choosing the slices of pizza, order is not important. This arrangement is a combination. Thus, we need to find C(4, 2) = 4! 2!(4−2)! = 6. So, there are six ways to choose two slices of pizza from the plate. Example 35.7 How many ways are there to select a committee to develop a discrete mathe- matics course at a school if the committee is to consist of 3 faculty members from the mathematics department and 4 from the computer science depart- ment, if there are 9 faculty members of the math department and 11 of the CS department? Solution. There are C(9, 3) · C(11, 4) = 9! 3!(9−3)! · 11! 4!(11−4)! = 27, 720 ways. Practice Problems Problem 35.11 Compute each of the following: (a) C(7,2) (b) C(8,8) (c) C(25,2) Problem 35.12 Find m and n so that C(m, n) = 13 5 Problem 35.13 The Library of Science Book Club offers three books from a list of 42. If you circle three choices from a list of 42 numbers on a postcard, how many possible choices are there? Problem 35.14 At the beginning of the second quarter of a mathematics class for elementary school teachers, each of the class’s 25 students shook hands with each of the other students exactly once. How many handshakes took place? Problem 35.15 There are five members of the math club. In how many ways can the two- person Social Committee be chosen? Problem 35.16 A consumer group plans to select 2 televisions from a shipment of 8 to check the picture quality. In how many ways can they choose 2 televisions? Problem 35.17 The Chess Club has six members. In how many ways (a) can all six members line up for a picture? (b) can they choose a president and a secretary? (c) can they choose three members to attend a regional tournament with no regard to order? Problem 35.18 Find the smallest value m and n such that C(m, n) = P (15, 2) Problem 35.19 A school has 30 teachers. In how many ways can the school choose 3 people to attend a national meeting? Problem 35.20 Which is usually greater the number of combinations of a set of objects or the number of permutations? Problem 35.21 How many different 12-person juries can be chosen from a pool of 20 juries? 6 Finding Probabilities Using Combinations and Permutations Combinations can be used in finding probabilities as illustrated in the next example. Example 35.8 Given a class of 12 girls and 10 boys. (a) In how many ways can a committee of five consisting of 3 girls and 2 boys be chosen? (b) What is the probability that a committee of five, chosen at random from the class, consists of three girls and two boys? (c) How many of the possible committees of five have no boys?(i.e. consists only of girls) (d) What is the probability that a committee of five, chosen at random from the class, consists only of girls? Solution. (a) First note that the order of the children in the committee does not matter. From 12 girls we can choose C(12, 3) different groups of three girls. From the 10 boys we can choose C(10, 2) different groups. Thus, by the Fundamental Principle of Counting the total number of committee is C(12, 3) · C(10, 2) = 12! 3!9! · 10! 2!8! = 12 · 11 · 10 3 · 2 · 1 · 10 · 9 2 · 1 = 9900 (b) The total number of committees of 5 is C(22, 5) = 26, 334. Using part (a), we find the probability that a committee of five will consist of 3 girls and 2 boys to be C(12, 3) · C(10, 2) C(22, 5) = 9900 26, 334 ≈ 0.3759. (c) The number of ways to choose 5 girls from the 12 girls in the class is C(10, 0) · C(12, 5) = C(12, 5) = 12 · 11 · 10 · 9 · 8 5 · 4 · 3 · 2 · 1 = 792 (d) The probability that a committee of five consists only of girls is C(12, 5) C(22, 5) = 792 26, 334 ≈ 0.03 7