...Combinations and Permutations What's the Difference? In English we use the word "combination" loosely, without thinking if the order of things is important. In other words: |[pic] |"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be | | |"bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. | | | | |[pic] |"The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be | | |exactly 4-7-2. | So, in Mathematics we use more precise language: |[pic] |If the order doesn't matter, it is a Combination. | |[pic] |If the order does matter it is a Permutation. | | |[pic] |So, we should really call this a "Permutation Lock"! | In other words: A Permutation is an ordered Combination. |[pic] |To help you to remember...
Words: 1778 - Pages: 8
...Combinations and Permutations What's the Difference? In English we use the word "combination" loosely, without thinking if the order of things is important. In other words: |[pic] |"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be | | |"bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. | | | | |[pic] |"The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be | | |exactly 4-7-2. | So, in Mathematics we use more precise language: |[pic] |If the order doesn't matter, it is a Combination. | |[pic] |If the order does matter it is a Permutation. | | |[pic] |So, we should really call this a "Permutation Lock"! | In other words: A Permutation is an ordered Combination. |[pic] |To help you to remember...
Words: 1778 - Pages: 8
...Permutations Permutations are a one to one correspondence of a set unto itself or a set that is arranged into a particular order. The number of permutations in a set can be counted. Counting these permutations can be easy if you have a small number of elements that only have one orientation. An example of this is if you have only three elements with one orientation then there would be 3! (Factorial) which is 3*2*1= 6 with the operation * being multiplication. If you take elements that have more than one orientation counting the permutations becomes exponential as the other orientations have to be counted also. If we take three elements with 2 orientations then we still have 3! but we have 3 elements with two orientations that will give...
Words: 1866 - Pages: 8
...Page |1 PERMUTATIONS and COMBINATIONS If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation. PRACTICE! Determine whether each of the following situations is a Combination or Permutation. 1. Creating an access code for a computer site using any 8 alphabet letters. 2. Determining how many different ways you can elect a Chairman and Co-Chairman of a committee if you have 10 people to choose from. 3. Voting to allow 10 new members to join a club when there are 25 that would like to join. 4. Finding different ways to arrange a line-up for batters on a baseball team. 5. Choosing 3 toppings for a pizza if there are 9 choices. Answers: 1. P 2. P 3. C 4. P 5. C Page |2 Combinations: Suppose that you can invite 3 friends to go with you to a concert. If you choose Jay, Ted, and Ken, then this is no different from choosing Ted, Ken, and Jay. The order that you choose the three names of your friends is not important. Hence, this is a Combination problem. Example Problem for Combination: Suppose that you can invite 3 friends to go with you to a concert. You have 5 friends that want to go, so you decide to write the 5 names on slips of paper and place them in a bowl. Then you randomly choose 3 names from the bowl. If the five people are Jay, Ted, Cal, Bob, and Ken, then write down all the possible ways that you could choose a group of 3 people. Here are all of the possible combinations of 3: Jay, Ted, Cal Jay, Ted, Bob Jay, Ted, Ken Jay, Cal...
Words: 1148 - Pages: 5
... Report page Share Share this 6.3 Probabilities Using Counting TechniquesThis is a featured page In a number of different situations, it is not easy to determine the outcomes of an event by counting them individually. Alternatively, counting techniques that involve permutations and combinations are helpful when calculating theoretical probabilities. This section will examine methods for determining theoretical probabilities of successive or multiple events. Permutation? or Combination? The following flow chart will help determine which formula is suitable for any given question. By simply following a series of "yes" or "no" questions, the appropriate formula can be determined. Flow Ex. 1 - Using Permutations: 6.3 Probabilities Using Counting Techniques - MDM4U1@FMG 6.3 Probabilities Using Counting Techniques - MDM4U1@FMG The specific outcome of Mike starting in lane 1 and the other two starting in lane 2 and lane 3 can only happen one way, so n(A) = 1. Therefore, 6.3 Probabilities Using Counting Techniques - MDM4U1@FMG The probability that Mike will start in the first lane next to his other brothers in lane 2 and 3 is approximately 0.00101. Ex. 1(a) - Using Permutations: Exactly Three People form a line at a grocery store. What is the probability that they will line up in descending order of age? (I.e. oldest, middle and youngest) →Solution using the blank like method: n(A): # of ways they will line up in descending order of age, thus: ...
Words: 1110 - Pages: 5
...700+ GMAT Problem Solving Probability and Combinations Questions With Explanations Collected by Bunuel Solutions by Bunuel gmatclub.com 1. Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his? A. 24/64 B. 32/64 C. 36/64 D. 40/64 E. 42/64 Expected value of a roll of one dice is 1/6(1+2+3+4+5+6)=3.5. Expected value of three dices is 3*3.5=10.5. Mary scored 10 so the probability to have more then 10, or more then average is the same as to have less than average=1/2. P=1/2. Answer: B. Discussed at: http://gmatclub.com/forum/mother-mary-comes-to-me-86407.html 2. Denise is trying to open a safe whose combination she does not know. IF the safe has 4000 possible combinations, and she can try 75 different possibilities, what is the probability that she does not pick the one correct combination. A. 1 B. 159/160 C. 157/160 D. 3/160 E. 0 When trying the first time the probability Denise doesn't pick the correct combination=3999/4000 Second time, as the total number of possible combinations reduced by one, not picking the right one would be 3998/3999. Third time 3997/3998 ... And the same 75 times. So we get: [pic] every denominator but the first will cancel out and every nominator but the last will cancel out as well. We'll get 3925/4000=157/160. Answer: C. Discussed at: http://gmatclub.com/forum/4000-possible-combination-84435...
Words: 7620 - Pages: 31
...permutation In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting (rearranging) objects or values. Informally, a permutation of a set of objects is an arrangement of those objects into a particular order. For example, there are six permutations of the set {1,2,3}, namely (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). One might define an anagram of a word as a permutation of its letters. The study of permutations in this sense generally belongs to the field of combinatorics. The number of permutations of n distinct objects is n×(n − 1)×(n − 2)×...×2×1, which number is called "n factorial" and written "n!". Permutations occur, in more or less prominent ways, in almost every domain of mathematics. They often arise when different orderings on certain finite sets are considered, possibly only because one wants to ignore such orderings and needs to know how many configurations are thus identified. For similar reasons permutations arise in the study of sorting algorithms in computer science. In algebra and particularly in group theory, a permutation of a set S is defined as a bijection from S to itself (i.e., a map S → S for which every element of S occurs exactly once as image value). This is related to the rearrangement of S in which each element s takes the place of the corresponding f(s). The collection of such permutations form a symmetric group. The key to its structure is the possibility...
Words: 440 - Pages: 2
...enumeration methods; sum rule, product rule, permutations, combinations along with enumeration methods for indistinguishable objects, how can we devise a strategy to solve problems requiring these methods? A basic concept in the branch of the theory of algorithms called enumeration theory, which investigates general properties of classes of objects numbered by arbitrary constructive objects (cf. Constructive object). Most often, natural numbers appear in the role of the constructive objects that serve as numbers of the elements of the classes in question ("Enumeration", 2013). The Sum/Difference Rules refer to the derivative of the sum of two functions is the sum of the derivatives of the two functions ("Basic Derivative Rules", 2013). The product rule is one of several rules used to find the derivative of a function. Specifically, it is used to find the derivative of the product of two functions. It is also called Leibnitz's Law, and it states that for two functions f and g their derivative (in Leibnitz notation, ). The derivative of f times g is not equal to the derivative of f times the derivative of g: .The product rule can be used with multiple functions and is used to derive the power rule. The product rule can also be applied to dot products and cross products of vector functions. The Leibnitz Identity, a generalization of the product rule, can be applied to find higher-order derivatives ("Definition Of Product Rule", 2013). A permutation in mathematics is one of several ways...
Words: 652 - Pages: 3
...Factorials, Permutations and Combinations Factorials A factorial is represented by the sign (!). When we encounter n! (known as 'n factorial') we say that a factorial is the product of all the whole numbers between 1 and n, where n must always be positive. For example 0! is a special case factorial. This is special because there are no positive numbers less than zero and we defined a factorial as a product of the numbers between n and 1. We say that 0! = 1 by claiming that the product of no numbers is 1. The reasoning and mathematics behind this is complicated and beyond the scope of this page, so let's just accept 0! as equal to 1. This works out to be mathematically true and allows us to redefine n! as follows: For example The above allows us to manipulate factorials and break them up, which is useful in combinations and permutations. Useful Factorial Properties The last two properties are important to remember. The factorial sign DOES NOT distribute across addition and subtraction. Permutations and Combinations Permutations and Combinations in mathematics both refer to different ways of arranging a given set of variables. Permutations are not strict when it comes to the order of things while Combinations are. For example; given the letters abc The Permutations are listed as follows Combinations on the other hand are considered different, all the above are considered the same since they have the exact same letters only arranged different. In other words, in combination...
Words: 895 - Pages: 4
...The principle of inclusion–exclusion will be used to count the permutations of n objects that leave no objects in their original positions. Consider Example 4. EXAMPLE 4 The Hatcheck Problem Anewemployee checks the hats of n people at a restaurant, forgetting to put claim check numbers on the hats. When customers return for their hats, the checker gives them back hats chosen at random from the remaining hats. What is the probability that no one receives the correct hat? ▲ Remark: The answer is the number of ways the hats can be arranged so that there is no hat in its original position divided by n!, the number of permutations of n hats.We will return to this example after we find the number of permutations of n objects that leave no objects in their original position. A derangement is a permutation of objects that leaves no object in its original position. To solve the problem posed in Example 4 we will need to determine the number of derangements of a set of n objects. EXAMPLE 5 The permutation 21453 is a derangement of 12345 because no number is left in its original position. However, 21543 is not a derangement of 12345, because this permutation leaves 4 fixed. ▲ Let Dn denote the number of derangements of n objects. For instance, D3 = 2, because the derangements of 123 are 231 and 312.We will evaluate Dn, for all positive integers n, using the principle of inclusion–exclusion. THEOREM 2 The number of derangements of a set with n elements is Dn = n! 1...
Words: 329 - Pages: 2
...Permutations Definition: A permutation of a set X is a rearrangement of its elements. Example: Let X = {1, 2, 3}. Then there are 6 permutations: 123, 132, 213, 231, 312, 321. Definition : A permutation of a set X is a one-one correspondence (a bijection) from X to itself. Notation: Let X = {1, 2, . . . , n} and α : X → X be a permutation. It is convenient to describe this function in the following way: α= 1 2 ... n α(1) α(2) . . . α(n) . Example: 1 2 2 1 1 2 3 1 2 3 1 2 3 2 3 1 1 2 3 4 1 4 3 2 1 2 3 4 5 3 5 4 1 2 Definition: Let X = {1, 2, . . . , n} and α : X → X be a permutation. Let i1 , i2 , . . . , ir be distinct numbers from {1, 2, . . . , n}. If α(i1 ) = i2 , α(i2 ) = i3 , . . . , α(ir−1 ) = ir , α(ir ) = i1 , and α(iν ) = iν for other numbers from {1, 2, . . . , n}, then α is called an r-cycle. Notation: An r-cycle is denoted by (i1 i2 . . . ir ). Example: 1 2 3 4 5 2 5 3 4 1 1 2 3 4 5 2 5 4 3 1 = (125) 3 − cycle is not a cycle 1 Remark: We can use different notations for the same cycles. For example, 1 2 3 1 2 3 = (1) = (2) = (3), 1 2 3 2 3 1 = (123) = (231) = (312). Warning: Do not confuse notations of a permutation and a cycle. For example, (123) = 123. Instead, (123) = 231 and 123 = (1). Composition (Product) Of Permutations Let α= Then α◦β = β◦α= 1 2 ... n α(β(1)) α(β(2)) . . . α(β(n)) 1 2 ... n β(α(1)) β(α(2)) . . . β(α(n)) , . 1 2 ... n α(1) α(2) . . . α(n) and β = 1 2 ... n β(1) β(2) . . . β(n) . Example: Let α = 1 2 3 4 5 5 1 2 4 3 α◦β...
Words: 1187 - Pages: 5
... The teacher will set up a debate as to who really invented Calculus. b. An essay quiz regarding the topic. Integration Method Integration is the process of unifying the child’s mental, physical, socio-emotional and spiritual energies through learning experience. Emphasis is placed on the development of integrated personality of the child. Learning activities are organized in terms of larger units. The teacher must also know her pupil’s like and dislikes. The child’s nature and experience must be made the starting points in planning and organizing school programs. Steps of the Integration Method. Subject: Probability Topic: Permutations and combinations 1. Introduction of the unit. Start off by explaining the objectives. After that, the teacher will present a pre-test about permutation that should be answered before the class ends. The teacher should correlate the lesson to the past lessons...
Words: 463 - Pages: 2
... T1 S2 T2 S3 Here n is 3, r is 2 For principle of counting we must have same number of options. In above if S3 and T2 is not allowed, principle of counting doesn’t work If n=5 and r=3 {A,B,C,D,E} How many different ways can we arrange of taking 3 letters at a time? 5 *4*3= 60 ways This is permutation of n different thing taken r at a time 60=(5*4*3*2*1)/(2*1) = 5!/2!=5!(5-3)!=n!/(n-r)! We are talking about linear arrangement not the circular one here nPr= filling r places by n different thing n=5 {A,B,C,D,E} r=3 {A,B,C}, {A,B,D}, {A,C,D}, {A,C,E}………….. [Note: Arrangement is related to permutation. If we are considered about place or position it is permutation question. Selecting is related to permutation. If we are not considered about place or position it is combination question.] nCr, C(n,r)= n!/[(n-r)!r!] n=5 {A,B,C,D,E} r=3 {A,B,C}, {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E},{A,D,E}, {B,C,D},{B,C,E} 3 are selected out of 5 . Above are possible combinations. Permutation: {A,B,C} and {A,C,B} are different because here, in {A,B,C} second place we have B but in {A,C,B} second place we have C Combination: {A,B,C} and {A,C,B} are same because in both of them letters A,B,C are selected....
Words: 399 - Pages: 2
...Unit 3- Review Organized Counting and Permutation [pic] 1. Some terms of a row of Pascal’s triangle are shown below. 1, 11, 55, 165, 330, 462, …… a. Determine the row number. b. Determine how many items are in this row. c. Determine the remaining terms in this row. d. Determine the sum of the numbers in this row. e. Determine the terms of previous row. 2. Express as a single term from Pascal’s triangle. a. [pic] b. [pic] c. [pic] d. [pic] e. [pic] f. [pic] 3. Express in factorial notation a. [pic] b. [pic] c. [pic] d. [pic] 4. Simplify. a) [pic] b) [pic] c) [pic] d) [pic] 5. Snack Shack serves: egg or ham sandwiches; coffee, soft drink or milk; donuts or pie for dessert. What are the possible meals if one item is chosen from each category? 6. Use the letters in the word " square " and tell how many 6-letter arrangements, with no repetitions, are possible if the : a. first letter is a vowel. b. vowels and consonants alternate, beginning with a consonant. 7. How many different 5-letter words can be formed from the word APPLE? 8. How many arrangements of the word ACTIVE are there if a. C,E must be together b. C,E must not be together 9. How many different six-digit numbers can be written using...
Words: 687 - Pages: 3
...Re-positioning Going from closed source to open source Challenges: 1. How to execute repositioning initiative? 2. Attracting developers from all around the world 3. Factors affecting the strategies 4. Competitor’s reaction 1. Repositioning * Starting a test run in the SAP organization itself to have a hand-on experiences * Identifying the areas of improvement * Integrate various modules and test all the permutations and combinations and develop logics accordingly * Creating a strong SOA * Compatible and easily upgradeable * Targeting the developing nations and encouraging medium scale companies to use it as net spending on technology by the companies have decreased. * SAP acquired * TomorrowNow * Manufacturer software Lighthammer to integrate shop floor operations * CommerceOne, for procurement. * Integrate all these software and make compatible with the SAP ERP * So ultimately creating a software which is more flexible and at lesser cost targeting SME’s and generating revenue from the developing nations * Creating awareness about the new product * Regular updates 2. Attracting developers * Form a common platform to work upon by the developers * Provide online training, circulating videos. * Asking for the feedbacks and inputs for the further improvement of the platform * Providing incentives to developers * Providing employment opportunities 3. Factors affecting the strategies * Integrating...
Words: 296 - Pages: 2
