1. PERMUTATIONSuppose n objects are to be ordered from 1st to nth, each order is called a permutation. Apply the multiplication principle to count the number of permutations of n objects, i.e. n(n-1)(n-2)(n-3)....(3)(2)(1), or n!, called n factorial.e.g. Suppose that 10 students are going on a bus trip, and each of the students will be assigned to one of the 10 available seats. What is the number of possible different seating arrangements of the students on the bus?Notice: n objects should be distinguishable and they are always ordered in a line. If the objects are not ordered in a line but in other shapes, such as a circle or a square, how to calculate the number of permutations?e.g. Five students are going to sit around a table, how many arrangement can there be? (If the relative position of two students is the same, then we view it as one arrangement.)formula: (n-1)!If there are some objects are exactly the same, the number of permutations should be calculated in another way.e.g. How many different five-letter words can be formed when all letters in the word ENTER are used each time.formula: n!/(number of repeated objects)!Suppose that k objects will be selected from a set of n objects, where k<=n, and the k objects will be placed in order from 1st to kth. The number of permutation is n(n-1)(n-2)....(n-k+1).e.g. How many different five-digit positive integers can be formed using the digits 1, 2, 3, 4, 5, 6, and 7 if none of the digits can occur more than once in the integer?2. CombinationGiven the five letters A, B, C, D, and E, determine the number of ways of selecting 3 of the 5 letters, but unlike before, you do not want to count different orders for the 3 letters. i.e.Note that n choose k is always equal to n choose n-ke.g. You should choose 3-person committee from a group of 9 students. How many ways are there to do this?The difference between permutation and combination is whether order is considered. If different order makes different arrangement, it is permutation, and if different order makes no difference, it is combination. Most problems require us to combine these two.e.g. Suppose you want to choose 5 members from a group of 8 to watering 5 different gardens. How many ways to do this?3. Permutation& combinationMultiplication Principle:If we need N steps to do a job and in the first step we have M1 different ways, second M2, third M3...., then we apply multiplication principle to count the total ways of doing this job, i.e. N=M1*M2*M3....*MnAddition Principle:If we have N different ways to do a job and in the first way we have M1 different ways, second M2, third M3..., then we apply addition principle to count the total ways of doing this job, i.e. N=M1+M2+M3...+MnNotice: In multiplication principle steps are interdependent, while in addition principle each way can independently make the job done.General method to solve permutation & combination problems:1. Understand what needs to be done2. Decide whether to take steps or divide into different ways, or both. When taking steps, how many steps are there; when dividing into different ways, how many ways in total.4. Make sure of the requirements in each step or way, permutation or combination, and calculate the number of arrangements in each step or each way.5. Count the total ways using multiplication principle or addition principle.3.1 Give priorities to special elements and placese.g. How many different 5-digit odd numbers can be formed when choosing from 0, 1, 2, 3, 4, 5 and each number can be used only once? (key: 288)3.2 Bundling Strategye.g. There are 7 students to stand in a straight line. If student A and B should always stand side by side, so should student D and G, how many ways to line those 7 students? (key: 480)Remember to make permutation between bundling elements3.3 Slot strategye.g. There is going to be 2 dances, 2 cross talks, and 1 solo in a party. If the 2 dances cannot be performed in a row, how many different program lists can we make for this party? (key: 72)Permutation first and then plug in the special elements3.4 More than one line problemse.g. Suppose you are going to arrange 8 students into two rows with four students in each row. Student A and B must be in the first row and Student C must in the second row. How many ways to arrange? (key:5760)Turn it into one straight line problem3.5 Mixture problemse.g. Suppose we need to put 5 different balls into 4 different boxes and each box should contain at least one ball. How many ways to make this done? (key: 240) First choose and then order3.6 Plank strategye.g. An elementary school has 10 athlete quota to be assigned to 7 classes. If every class should have at least one quota, how many ways are there to assign? (key: 84)Pay attention to the differences between 3.5 and 3.64. ProbabilityProbability is a way of describing uncertainty in numerical terms.Probability experiment (random experiment)The set of all possible outcomes of a random experiment is called the sample space, and any particular set of outcomes is called an event.P(E)=the number of outcomes in the event/the number of possible outcomes in the experimentGeneral facts about probability:●If an event E is certain to occur, then P(E)=1●If an even E is certain not to occur, then P(E)=0●If an even E is possible but not certain to occur, then 0<P(E)<1●The probability that an event E will not occur is equal to 1-P(E)●If E is an event, then the probability of E is the sum of the probabilities ofoutcomes in E.●The sum of the probabilities of all possible outcomes of an experiment is 1. If E and F are two events of an experiment, we consider two other events relatedto E and F.●The event that both E and F occur, that is, all outcomes in the set●The event that E or F, or both, occur, that is , all outcomes in the set●P(E or F)=P(E)+P(F)-P(both E and F)●P(E or F)=P(E)+p(F) if E and F are mutually exclusive●P(E and F)=P(E)P(F) if E and F are independent●the probability of one event occurring k times during n independentexperiments P(E)=e.g. A fair 6-sided die is rolled nine times and what's the probability that two of the nine are odd number?If a fair 6-sided die is rolled once, let E be the event of rolling a 3 and let F be the event of rolling an odd number. What is the probability of both E and F occurring?A 12-sided die, with faces numbered 1 to 12, is to be rolled once, and each of the12 possible outcomes is equally likely to occur. What is the probability of rollinga number that is either a multiple of 5 or an odd number?Consider an experiment with events A, B, and C for which P(A)=0.23, P(B)=0.40, and P(C)=0.85. Suppose that events A and B are mutually exclusive and events B and C are independent. What are the probabilities P(A or B) and P(B or C)?Suppose that there is a 6-sided die that is weighted in such a way that each time the die is rolled, the probabilities of rolling any of the numbers from 1 to 5 are all equal, but the probability of rolling a 6 is twice the probability of rolling a 1. When you roll the die once, the 6 outcomes are not equally likely. What are the probabilities of the 6 outcomes?Suppose that you roll the weighted 6-sided die from above example twice. What is the probability that the first roll will be an odd number and the second roll will be an even number?A box contains 5 orange disks, 4 red disks, and 1 blue disk. You are to select two disks at random and without replacement from the box. What is the probability that the first disk you select will be red and the second disk you select will be orange?There are going to be 5 different songs and 3 different dances in a particular party. How many ways to arrange the programs as follows?(1) three dances should be shown in a row(2) three dances should be separatedHow many ways to divide 6 different books into three piles?How many ways to assign 6 different books to 3 persons and each has exactly two books?How many different 5-digit even numbers can be formed from 0, 2, 3, 6, and 9? (Each number should be used only once.)A. 120B. 72C. 60D, 48E. 106Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?(A)11/8(B)7/8(C)9/64(D)5/64(E)3/64Among a group of 2,500 people, 35 percent invest inmunicipal bonds, 18 percent invest in oil stocks, and7 percent invest in both municipal bonds and oilstocks. If 1 person is to be randomly selected fromthe 2,500 people, what is the probabilitythat theperson selected will be one who invests in municipalbonds but NOT in oil stocks?A certain club has 10 members, including Harry. Oneof the 10 members is to be chosen at random to bethe president, one of the remaining 9 members is tobe chosen at random to be the secretary, and one ofthe remaining 8 members is to be chosen at randomto be the treasurer. What is the probability that Harrywill be either the memberchosento be the secretaryor the member chosen to be the treasurer?A={2, 3, 4, 5}, B={4, 5, 6, 7, 8}Two integers will be randomly selected from the setsabove, one integer from set Aand one integer fromset B. What is the probability that the sum of the twointegers will equal 9 ?。