Math141_Fall_2012 ( Business Mathematics 1) Week 7. Dr. Marco A. Roque Sol Department of Mathematics Texas A&M University


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1 ( Business Mathematics 1) Week 7 Dr. Marco A. Roque Department of Mathematics Texas A&M University
2 In this sections we will consider two types of arrangements, namely, permutations and combinations a. each permutation consists of three letters a, b, c. Thus, we may think such a sequence as being constructed by filling in each of the three blanks Definition Given a set of distinct objects, a permutation of the set is an arrangement of these objects in a definite order. That the order is important, can be seen from the situation where we have to deal with University Student Identification Numbers, so it is not the same to have the number than even though we have the same digits. 1st letter 2nd letter 3rd letter To pick the first letter we have three options, once the first letter is picked, for the second letter we just have two options and finally after the first two letters have been selected, then for the third letter we just have one option. Thus, by the multiplication principle we have Example Let A = { a, b, c } a. Find the number of permutations of A. b. List all the permutations of A with the aid of a tree diagram That is, we have (3)(2)(1) = 6 possible permutations
3 b. Using a tree diagram For the first place we have nine options, then for the second place we have eight options,.., for the ninth place we just have one option. Then, by the multiplication principle we have (9)(8)(7) (1) = 362, 880 Definition 2) Find the number of ways in which a baseball team consisting of nine people can arrange themselves in a line for a group picture. For any natural number n, the factorial of n denoted by n! is defined as n! = n (n1)(n2) 1 OBS In this case we have nine places to fill in. n! = n(n1)! n! = n(n1)(n2)!
4 Using the factorial of a number, we can give a way to calculate the number of permutations of a set with n members. In this way, the number of permutations of n distinct objects taken n at a time, denoted by P(n,n) is given by P(n,n) = n! Therefore by the multiplication principle, there are (n) (n1) (n2) (nr+1) different permutations of n elements taken r at a time, but the above quantity is equivalent to Now, in some cases, given a set of n distinct objects, we would be interested in finding the number of permutations of the set taken r ( n) at a time. In that situation we will have r different places to be fill in (n ) (n 1 )... (n r+1 ) (n r ) (n r 1 ) (n r 2 )...(1) (n ) (n 1 )... (n r+1 )= (n r ) (n r 1 ) (n r 2)... (1 ) (n ) (n 1 )... (n r+ 1 )= n! (n r )! Therefore r Permutations of n distinct objects Thus, the first place has n options, then the second places has (n1) options, the third place has (n2),.., and the rth place has ( nr+1) options. The number of permutations of n distinct objects taken r at a time, denoted by P(n,r), is given by P (n,r ) = n! (n r )!
5 P (8,4) = 8! (8 4 )! = 8! 4! = 1680 Examples 1) Let A = { a, b, c, d}. find all the permutations of A taken two at a time. Here, n = 4 and r = 2, so 2) Find the number of ways a President,Vicepresident, Secretary, and a Treasurer can be chosen from a committee of eight members. P (4,2 ) = 4! (4 2 )! = 4! 2! = 12 Since the order is important, we have to find the permutations of 8 distinct objects taken 4 at a time So far, we have considered permutations of n distinct objects taken r at a time, but what if we have inside of the set, some subgroups of identical elements? In that case we have to take out all the permutaions of the identical subgroups. Example Find all thedistinct permutations of the word ATLANTA. The word consists of 7 letters, and the number of permutations is given by 7!, but we have to consider that there are two subgroups of identical elements. The first one associated to the letter A with three members and the second one associated to the letter T with two members, that is, we need to take the permutations of each subgroup out of the final result. Therefore, the total number of permutations is have P (4,2 ) = 7! 3!2! = 420
6 In general we obtain the folowing result Permutations of n Objects, Not all Distincts Given a set of n objects in which n 1 are alike and of one kind, n 2 are alike and of another kind,, and n m are alike and of yet another kind, so that n 1 + n n m = n Then the number of permutations of these objects taken n at a time are given by Example n! n 1!n 2!...n m! If nine students are taking a test and they will be graded by three instructors, taking each one three exams. In how many ways can the tests be directed to any of the three instructors? We can think in the nine tests as an arrange of nine places, and each one is going to be associated to one of the instructors, but each instructor receives three tests, then the problem is equivalent to find the permutions of nine elements with three subgroups of identical elements.thus, the total number of ways the nine test can be distributed to the instructors is given by 9! 3!3!3! = 1680 Now, when the order is not important in the arrangement, that is, when we want to select r objects from a set of n objects without any regard to the order in which these objects are selected, then we are talking about Combinations. In this way to find the number of combinations de n elements taken r at a time, from the number of permutations P (n,r ) = n! (n r )!
7 We have to take out the permutations of each arrangement of r elements, that is, Combinations of n objects The number of combinations of n distinct objects taken r at a time is given by OBS C(n,r) gives us the number of subsets with r ( n) elements from a set with n elements. Examples C (n,r ) = P (n,r ) r! 1) How many poker hands of 5 cards can be dealt from a standerd deck of 52 cards? = n! r! (n r )! n! r! (n r )! ( where r n ) Since the order does not matter, we just want to find the combinations of 52 distinct elements taken 5 at a time. 2) A Senate investigation subcommittee of four members is to be selected from a Senate committee of ten members. Determine the number of ways in which this can be done. C (52,5 ) = 52! 5! (52 5 )! = 52! 5! 47! = 2,598,960 Since in the selection of the members the order does not matter, we need to find the combinations of 10 members taken 4 at a time. C (10,4 ) = 10! 4! (10 4 )! = 10! 4!6! = 210
8 3) The members of a string quartet consisting of two violinists, a violist, and a cellist are to be selected from a group of six violinists, three violists, and two cellists. a. In how many ways can the string quartet be formed? b. In how many ways can the string quartet be formed if one of the violinists is to be designed as the first violinist and the other is to be designated as the second violinist? b. In this case the order is important for the violinists. The violinists can be selected in P(6,2) ways The violists can be selected in C(3,1) ways The cellists can be selected in C(2,1) ways and by the multiplication principle we have P(6,2) C(3,2) C(2,1) = (30)(3)(2) = 180 a. The violinists can be selected in C(6,2) ways The violists can be selected in C(3,1) ways The cellists can be selected in C(2,1) ways 4) The Psicos, a pop group, are planning a concert tour with performances to be given in five cities: San Francisco, Los Angeles, San Diego, Denver, and Las Vegas. In how many ways can they arrange their itinerary if and by the multiplication principle we have C(6,2) C(3,1) C(2,1) = (15)(3)(2) = 90 a. There are no restricions. b. The three performances in California must be given consecutively?
9 In both cases the order is important a. P(5,5) = 5! = 120 b. Now the three cities of california can be thought as a block and the other cities as two more blocks. Thus, the number of ways of performing in California and the other cities is P(3,3). On the other hand inside of the California block, we have P(3,3) ways to arrange the itinerary. Therefore, using the multiplication principle we have. (3!)(3!) = 36 There are : C(5,2) ways the investor can select the aerospace companies C(3,2) ways the investor can select the energy companies C(4,2) ways the investor can select the electronics companies Therefore by the multiplication principle, there are C(5,2)C(3,2)C(4,2) = 180 5) Suppose that an investor has decided to purchase shares in the stocks of two aerospace companies, two energy development companies,and two electronic companies. In how many ways can the investor select the group of six companies from the investment from the recommended list of five aerospace companies, three energy development companies, and four electronics companies?
10 Probability. Experiments, Sample Space, and Events Sample Point, Smaple Space, and Event There are many terms used in the probability theory and we will start giving some basic definitions. Definition An experiment is an activity with observable results. Example 1)Tossing a coin and observing whether it falls heads or tails 2) Rolling a die and looking the number falling uppermost 3) Testing a light bulb from a batch and observing whether is deffective or not In the discussion of expermients, we use the following terms: Sample Point: An outcome of an experiment. Sample Space: The set consisting of all possible sample points of an experiment. Event: A subset of a sample space of an experiment. From the above set of definitions we can see that the sample space plays the role of a universal set, that is, it will be our reference set in the study of probability. In this sense the sample space represents the certain event, that is, the event that must occurs because it contains all the possible outacomes of the experiment. On the other hand, the empty set Ø is called the impossible event, since the empty set contains no elements.
11 Probability. Experiments, Sample Space, and Events Example Uniond of Two events The union of two events E and F is the event E F Describe the sample space associated with the experiment of tossing a coin and observing whether it falls heads or tails. What are the events of this experiment? The sample space is given by S = { H, T } And all the possible subsets ( events ) are Ø, {H}, {T}, S Now, since every event is a subset of the sample space, then we can form new events using the set operations defined before. Thus, if E and F are events associated to an experiment with sample space S then we define the : Intersection of Two Events The instersection of two events is the event E F Complement of an Event The complement of the event E is the evemt E c Thus, Ven diagrams will be useful in solving problems dealing with events associated to some given experiment. Definition The events E and F are called mutually exclusive if E F = Ø
12 Probability. Experiments, Sample Space, and Events Examples 1) An experiment consists of tossing a coin three times and observing the resulting sequence of heads and tails a. Describe the sample space of the experiment. b. Determine the event E that exactly two heads appear. c. Determine the event F that at least one head appears. a. Since every time the coin has two different ways of landing, Head or Tail, then we have: S= { HHH,HHT,HTH,THH,HTT,THT,TTH, TTT} b. E ={HHT, HTH, THH} c. F = { HHH,HHT,HTH,THH,HTT,THT,TTH } 2) An experiment consist of rolling a pair of dice and observing the number that falls uppermost on each die. a. Describe an appropriate sample space for this experiment. b. Determine the events that E 1,E 2,, E 12, that the sum of the numbers falling uppermost is 1, 2,, 12. a. Since the first die has 6 possible otpions and the second die has also 6 possibilities, then by the multiplication principle we have 36 possible outcomes: S = { (1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2,1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3,1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5,1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6,1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
13 Probability. Experiments, Sample Space, and Events Examples b. E 1 = Ø E 2 = {(1, 1)} E 3 = {(1, 2), (2,1)} E 4 = {(1, 3), (2, 2), (3, 1)} E 5 = {(1, 4), (2, 3), (3, 2), (4, 1)} E 6 = {(1, 5),(2, 4). (3, 3), (4, 2), (5, 1)} E 7 = {(1, 6), (2,5), (3, 4), (4, 3), (5, 2), ( 6, 1)} E 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) } E 9 = {(3, 6), (4, 5), (5, 4), (6, 3) } 3) An experiment consist of recording, in order of their births, the sex composition of a threechild family in which the children were born at different times. a. Describe an appropriate sample space S for this experiment. b. Describe the event E that there are two girls and a boy in the family c. Describe the event F That the oldest child is a girl d. Describe the event G that the oldest child is a girl and the yougest child is a boy. a. Using a tree diagram E 10 = {(4, 6), (5, 5), (6, 4) } E 11 = {5, 6), (6,5)} E 12 = {(6, 6)}
14 Probability. Experiments, Sample Space, and Events Examples The sample space is given by S = { bbb, bbg, bgb, gbb, bgg,gbg,ggb, ggg } b. E = {bgg, gbg, ggb} c. F = { gbb, gbg, ggb, ggg} d. G = { gbb, ggb}
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