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Chapter 11:Permutations, Combinations and Binomial TheoremLessons 1 and 2:PermutationsSpecific Outcome 1: Apply the fundamental counting principle to solve problems.Achievement Indicator:1.1 Count the total number of possible choices that can be made, using graphic organizers such as lists and tree diagrams1.2 Explain, using examples, why the total number of possible choices is found by multiplying rather than adding the number of ways the individual choices can be made.1.3 Solve a simple counting problem by applying the fundamental counting principle.Specific Outcome 2: Determine the number of permutations of n elements taken r at a time to solve problems.Achievement Indicator:2.1 Count, using graphic organizers such as lists and tree diagrams, the number of ways of arranging the elements of a set in a row.2.2 Determine, in factorial notation, the number of permutations of n different elements taken n at a time to solve a problem.2.3 Determine, using a variety of strategies, the number of permutations of n different elements taken r at a time to solve a problem.2.4 Explain why n must be greater than or equal to r in the notation nP r .2.5 Solve an equation that involves nP r notation, such as nP 2 =302.6 Explain, using examples, the effect on the total number of permutations when two or more elements are identical.Fundamental Counting Principle:Ex.A toy manufacturer makes a wooden toy in three parts: Part 1:the top part may be coloured red, white or blue Part 2:the middle part may be orange or black, Part 3:the bottom part may be yellow, green, pink, or purpleDetermine how many different coloured toys can be produced:3.2. 4 = 24topmiddlebottom coloured toysCheck your answer using a tree diagram (if they don’t know how to make one, show them)Make a conjecture about how you can use multiplication only to arrive at the number of different coloured toys possible.The Fundamental Counting PrincipleConsider a task made up of several stages.If the number of choices for the first stage is a, the number of choices for the second stage is b, the number of choices for the third stage is c, etc., then the number of ways in which a task can be completed is a x b x c x …..This is called the fundamental counting principle.Another thing that you will need to consider is the idea of and vs. or.If the problem has “and” in it, that means you are to multiply.If the problem has “or” or “at least” or “at most”, it is talking about multiple situations and you need to add.Ex.Determine the number of distinguishable four letter arrangements thatcan be formed from the word ENGLISH if: a.letters can be repeated? 7. 7.7 .7= 2401no letters are repeated and:there are no further restrictions? 7 .6.5.4= 840the first letter must be E? 1. 6 .5.4= 120 Ethe “word” must contain G? 1 .6.5.4= 120(120)(4) – the ‘G’ can be in any of the four slots= 480the first and last letters must be vowels? 2_.5 .4 .1= 40 VVEx.The telephone numbers allocated to subscribers in a rural area consist of one of the following:the digits 345 followed by any three further digits orthe digit 2 followed by one of the digits 1 to 5, followed by any three further digits.How many different telephone numbers are possible?1.1 . 1.10.10 . 10or1 . 5. 10 . 10. 10= 1000+5000= 6000 different telephone numbers Ex. Car number plates in an African country consist of a letter other than I or O followed by three digits, the first of which cannot be zero, followed by any two letters which are not repeated.How many different car number plates can be produced? 24 . 9.10.10.26.25=14 040 000Ex.Consider the digits 2, 3, 5, 6, 7 and 9.If repetitions are not permitted, how many 3-digit number can be formed? 6.5.4= 120How many of these are:i)less than 400?2.5.4= 40ii)even?5.4.2= 40( only 5 choices because between the 2 and the 6, only one will be available for the first number because one of them is needed for the last, to be even)iii)multiples of 5? 5.4.1= 20 ( the last digit can only be 5, so that leaves 5 choices for the first digit)Factorial NotationConsider how many ways there are of arranging 6 different books side by side on a shelf.In this example we have to calculate the product 6x5x4x3x2x1.In mathematics this product is denoted by 6! (“factorial” or “factorial 6”)In general n!=n(n-1)(n-2)(n-3)….(3)(2)(1), where Ex.Use the factorial key on your calculator to solve 6!On Calc: Math – PRB – 4 - != 720Ex.To simplify, there are several approaches:a.Use you calculator: = 720 b. By Cancellation:10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1= 7207 x 6 x 5 x 4 x 3 x 2 x 1Ex.Find the value of Ex.Simplify the following expressions:PermutationsAn arrangement of a set of objects in which the order of the objects is important is called a permutation.Ex.How many permutations are there of the letters of the word: a.REGINA:6! = 720(6 letters)b.KELOWNA:7! = 5040(7 Letters) The number of permutations of “n” different objects taken “r” at a time is:(no repeats are allowed in this formula)Ex.Use thekey on your calculator to evaluate . Then solve usingfactorials.Using Factorials:Defining 0!If we replace r by n in the above formula we get the number of permutations of n objects taken n at a time. This we know is n!For this to be equal to n! , the value of 0! must be 1.Therefore 0! Is defined to have a value of 1Ex.In a South American country, vehicle license plates consist of any 2 different letters followed by 4 different digits.Find how many different license plates are possible using:a) fundamental counting principle: 26 .25.10.9. 8 .7 = 3 276 000b) permutations:(letters)(digits)Ex.Solve for n in the equation n!(n-3)! = 28(n-1)!(n-4)! (cross multiplied)n(n-1)!(n-3)! = 28(n-1)!(n-4)!(n is expanded to n(n-1)! to simplify)n(n-3)! = 28(n-4)!(cancel n-1 on each side)n(n-3)(n-4)!=28(n-4)! (expand n(n-3)!)n(n-3) = 28(cancel n-4’s)(n-7)(n+4) = 0n=7 and n = -4 (extraneous – doesn’t fit equation)In many cases involving simple permutations, the fundamental counting principle can be used in place of the permutation formulas.Permutations With Restrictions And RepetitionIn many problems restrictions are placed on the order in which objects are arranged.In this type of situation deal with the restrictions first.Ex.In how many ways can all of the letters of the word ORANGES bearranged if:a.)there are no further restrictions:7! = 5040 b.)the first letter must be an N? : 1 .6.5.4.3.2.1= 720Ex.Find the number of permutations of the letters in the word KITCHEN if: a.)the letters K, C, and N must be together but not necessarily in that order. The letters must be considered as a group of 1.Therefore, you are arranging that one group plus 4 other letters.So 5!.However, there you need to arrange the three letters themselves which is 3!.So (5!)(3!) = 720b) the letters IT cannot be together:It is easier to solve these questions if you consider the opposite situation Must Be together.That way, you can take the total with no restrictions and then subtract out the Must Be Together.The leftovers will be the must not be together answer.Total no restrictions:7! = 5040Must be together:IT = 1 + 5 other letters = 6! X 2! (arrangement of IT) = 1440Answer:5040 – 1440 = 3600Ex.In how many different ways can 3 girls and 4 boys be arranged in a row if no two people of the same gender can sit together?4.3.3.2.2.1.1 =144BGB GB GBPermutations with RepetitionsThe following formula gives the number of permutations when there are repetitions:The number of permutations of n objects, where a are the same of one type, b are the same of another type, and c are the same of yet another type, can be represented by the expression below**basically remember to divide out the repeats in factorial notation**Ex.Find the number of permutations of the letters of the word: a.) VANCOUVER: there are 2V’s:MATHEMATICAL:2M’s, 3A’s, 2T’s:Ex.How many arrangements of the word POPPIES can be made undereach of the following conditions?without restrictions:if each arrangement begins with a P:POPPIES:The first letter is a P, and the next 6 letters can be in any arrangement (remember there is a repetition of 2P’s within the remaining 6 letters)=Ex.Brett bought a carton containing 10 mini boxes of cereal.There are 3 boxes of Corn Flakes, 2 boxes of Rice Krispies, 1box of Coco Pops, 1box of Shreddies, and the remainder are Raisin Bran.Over a ten day period Brett plans to eat the contents of one box of cereal each morning.How many different order are possible if on the first day he has Raisin Bran?CF, 2 RK, 1 CP, 1 S, 3 RB (will only have 2 after first day)= One thing that you may see on the diploma are what we call pathway questions.They can be done by using repeated permutations or drawing.Consider the following problem:Ex.A city centre has a rectangular road system with 5 streets running north to south and 6 avenues running west to east.a.Draw a grid to represent this situation Sean is driving a car and is situated at the extreme northwest corner of the city centre.In how many ways can he drive to the extreme southeast corner if at each turn he moves closer to his destination (assume all streets and avenues allow two –way traffic)4N 5W = = 126 ways(also show them how to draw these)Assignment: pg. 524-527 #1 – 8, 10 – 20, 22 – 25, 27 – 28, 32, C3a, C5 (pick one letter from each question)Lesson 3: CombinationsSpecific Outcome 3: Determine the number of combinations of n different elements taken r at a time to solve problems.Achievement Indicator:3.1 Explain, using examples, the difference between a permutation and a combination.3.2 Determine the number of ways that a subset of k elements can be selected from a set of n different elements.3.3 Determine the number of combinations of n different elements taken r at a time to solve a problem.3.4 Explain why n must be greater than or equal to r in the notation nCr3.5 Explain, using examples, why n Cr =n C n- r3.6 Solve an equation that involves n C r such asn C 2 = 15When is order not important?1.From a group of four students, three are to be elected to an executive committee with a specific position.The positions are as follows:1st position President2nd position Vice President3rdposition Treasurer a.Does the order in which the students are elected matter?Why?Yes.The vice president is not the president (and vice versa).They have distinct roles. b.In how many ways can the positions be filled from this group?4 x 3 x 2 = 242.Now suppose that the three students are to be selected to serve on a committee.a.Is the order in which the three students are selected still important?Why or why not?Not important, there are no distinct titles.b.How many possible committees from the group of four students are now possible? 41 (President, VP, Treasurer),2(President, VP, student 4), 3( VP, Treasurer, student 4), 4( President, Treasure, student 4)3.You are part of a group of 6 students.a.How many handshakes are possible if each student shakes every other student’s hand once?5 + 4 + 3 + 2 + 1 = 154.What is the biggest difference between a permutation and a combination? A combination is a selection of a group of objects, taken from a larger group for which the kind of objects selected is important, but not the order in which they are selected.There are several ways to find the number of possible combination.One is to use reasoning.Use the fundamental counting principle and divide by the number of ways that the object can be arranged among themselves.For example, calculate the number of combinations of three digits made from the digits 1, 2, 3, 4, and 5: 5 x 4 x 3 = 60However, 3 digits can be arranged 3! ways among themselves. So:603!=10The number of combinations of “n” items take “r” at a time is:Formula:Use the formula to solve the last problem:5!5-3!3!=10Thekey on the calculator can be used to evaluate combinations:In some textsis written as Ex.Three students from a class of 10 are to be chosen to go on a school trip.In how many ways can they be selected? Ex.To win the LOTTO 649 a person must correctly choose six numbers from 1 to 49.Jasper, wanting to play LOTTO 649, began to wonder how many numbers he could make up.How many choices would Jasper have to make to ensure he had the six winning numbers?Ex.The Athletic Council decides to form a sub-committee of seven council members to look at how funds raised should be spent on sports activities in the school.There are a total of 15 athletic council members, 9 males and 6 females.The sub-committee must consist of exactly 3 females. a.In how many ways can the females be chosen? b.In how many ways can the males be chosen? c.In how many ways can the sub-committee be chosen?Exactly 3 females(females)(Males) d. In how many ways can the sub-committee be chosen if Bruce the football coach must be included?(females)(males)(Bruce)Combinations which are equivalentEx.Jane calculatedto be 45 arrangements.She then calculatedto be 45 arrangements.Use factorial notation to prove this: 45 = 45Solving for “n” in Combinations ProblemsEx.During a Pee Wee hockey tryout, all the players met on the ice after thelast practice and shook hands with each other.How many playersattended the tryouts if there were 300 handshakes in all?A handshake requires 2 people 25 players attended the tryoutsHomework:pg. 534#1 – 6, 7a, 8 – 20, 22, 23, C1, C3 (pick one letter from each question)#12 is a challengeLesson 4: The Binomial TheoremSpecific Outcome 4:Expand powers of a binomial in a variety of ways, including using the binomial theorem (restricted to exponents that are natural numbers).Achievement Indicators:4.1 Explain the patterns found in the expanded form of (x + y), n ? 4 and n?N, by multiplying n factors of (x + y).4.2 Explain how to determine the subsequent row in Pascal’s triangle, given any row.4.3 Relate the coefficients of the terms in the expansion of (x + y) to the (n + 1) row in Pascal’s triangle.4.4 Explain, using examples, how the coefficients of the terms in the expansion of (x + y)are determined by combinations.4.5 Expand, using the binomial theorem, (x + y)4.6 Determine a specific term in the expansion of (x + y)Pascal’s Triangle:Show the first couple rows of the Pascal’s triangle.See if the students can find the pattern and continue the next few rows.Ex.Complete the following expansions: a.x+y2x2+2xy+y2 b.x+y3x3+3x2y+3xy2+y3 c.x+y4x4+4x3y+6x2y2+4xy3+y4How do the coefficients of the simplified terms in your binomial expansions relate to Pascal’s Triangle?The coefficients related to the row of pascal’s.Pascal’s Triangle can also be represented using combinations:These values are identical to the values in Pascal’s triangle above.Notice these results from Pascal’s Triangle:the sum of the numbers in the kth row of Pascal’s triangle is take the 3rd row, add it up (equals 4), the sum of the coefficients in the expansion ofis(notice the number of terms (4))coefficients = 1 + 3 + 3 + 1 = 8()Notice these results from combinatorics:1 +6+15 + 20+ 15+6 + 1= 64 (pascal’s 7th row)Binomial Theorem, where All binomial expressions will be written in descending order of the exponent of the first term in the binomial.The following are some important observations about the expansion of , where x and y represent the terms of the binomial and n?N:The expansion contains n + 1 termsThe sum of the exponents in any term of the expansion is n.Ex.Expand 4 terms, n = 3, x = 3x, y = (-2)General Term of the Expansion of The termis called the general termof the expansion. It is theterm in the expansion (not term k)Ex.a)Find the fifth term of 9 terms, n=8, k = 4b) the middle term of 7 terms, n = 6, k = 3 (middle term would be the 4th term)Ex.One term in the expansion ofis 3 281 250 .Determine the numerical value of a.n = 10 , k = (10 – 4(exponent) = 6) this gives us the same x value power, the position of x4 in the expansion.The k value of 6 means the 7th term (7 – 1 = 6)y0+y1+y2+y3+y4+y5+y6 (this represents the k value of 6 for the y in (x +y)n, notice it is 7 terms)x10+x9+x8+x7+x6+x5+x4 (expansion for the x in (x +y)n, notice it is 7 terms)This ensures the x values will cancel out because they will have the same exponent.Ex.Find the constant term (the term independent of x) in the expansion of 2x-1x215General Strategy:1.Put equation into general term formula2.Simplify/ expand formula3.Solve for exponent of variables (gives k value)Rewrite equation : 2x+-1x215,n = 15, k = k=nCk xn-kyk=15Ck 2x15-k-1x2k=(15Ck) 215-kx15-k-1k1x2k=(15Ck) -1kx15-k215-kx-2k=(15Ck) -1kx15-kx-2k215-k=(15Ck) -1kx15-3k215-kx15-3k=by definition the exponent will equal 015-3k=05=k, tells us the term numberUse general term formula:=15C5 2x10-1x25=30031024x10-1x10=-3075072Assignment p. 542#1 – 11, 13 – 20, 22 – 23, C2 (pick one letter from each question)
