MA40S PRECALCULUS PERMUTATIONS AND COMBINATIONS CLASS NOTES

Transcription

1 1 MA40S PRECALCULUS PERMUTATIONS AND COMBINATIONS CLASS NOTES LESSON 1 The Fundamental Counting Principle Objective: To develop the Fundamental Counting Principle (FCP). Investigate Counting without Counting 1. A café has a lunch special consisting of an egg or a ham sandwich (E or H); milk, juice, or coffee (M, J, or C); and yogurt or pie for dessert (Y or P). a) One item is chosen from each category. List all possible meals. (List them: eg: EMY for Egg Milk Yogurt) b) How many possible meals are there? c) How can you determine the number of possible meals without listing all of them? (Make an outcome tree) MA40SP_E_PermsCombsNotes.doc Revised:

2 2 2. This café also features ice cream in 24 flavours. You can order regular, sugar or waffle cones. Suppose you order a double cone with two scoops of ice cream. a) How many choices of type of cone do you have? b) How many choices of flavour for the first scoop of ice cream do you have? c) How many choices do you have for the second scoop (you are allowed two scoops the same assume)? d) How many different double cones are possible? e) Would you have wanted to make an outcome tree for this problem?? Example 1: 3. A computer store sells 2 different computers, 3 different monitors, and 2 different printers. How many different computer systems are available? Outcome Tree. Drawing an outcome tree can often help also: The Fundamental Counting Principle If one item can be selected in m number ways, and for each of those ways a second item can be selected in n ways, then the two items can be selected in m n ways. This can be extended to any number of items. Eg: if you have three choices of shoes, 4 choices of skirts, and 2 choices of tops you have 342 = 24 choices of outfit to ear to school.

3 3 Example 2 Arrangements with restrictions: 4. How many different 2-digit numbers are there? (assume we do not write numbers with a leading zero ) Example 3 - Arrangements with restrictions: 5. In each case below, how many 2-digit numbers can be formed using the digits 0, 1, 3, 5, 7 and 9? (assume again we do not start numbers with a leading zero) a) Repetitions are allowed. b) Repetitions are not allowed. Repetition or replacement of items gives a greater number of possible arrangements than no replacement or no repetition. Example 4: 6. A true-false test has eight questions. Suppose students answer each question by randomly guessing. a) How many possible answers are there for each question? b) How many different arrangements of answers are possible for the answers to the eight questions on the test? c) What is the probability (or chance) that a student answers all eight questions correctly? We learn more about Probability in a later unit.

4 4 Example 5: 7. A multiple-choice test has eight questions, with four possible answers for each question. Suppose students answer each question by guessing randomly. a) How many possible answers are there for each question? b) How many different arrangements of answers are possible for the answers to the eight questions on the test? c) What is the probability or chance that all eight questions will be answered correctly? (we will study probability more fully in a different unit) 8. Example. My wife and I like to make local road trips in the summer. We like to go from Winnipeg to Brandon and then to Dauphin then back to Brandon and then return to Winnipeg. But we do not ever take the same road twice between towns thereby seeing more prairie scenery. If there are four roads connecting Winnipeg to Brandon and three connecting Brandon to Dauphin, how many different road trips can we take if we don t use the same road twice? Dauphin Brandon Winnipeg 9. Example. How many possible license plates are there for Manitoba? (Three letters followed by three single digits; assume no vanity plates) ( L L L # # # )

5 5 10. Example. A bag contains 15 marbles: blue and red. You notice that only one of the marbles is red; so how many are blue? (sometimes the easiest way to count what you are looking for is to count what you are not looking for (or the rest or the complement ) and subtract from the total!) number(blue) or n(blue) or n blue = 11. Example. How many license plates in Manitoba have at least two number digits the same? Hint: At least two means more than one. More than one the same means some repeats; so the complement of no repeats. a. how many have no two digits the same? b. how many total are possible? c. So how many have at least two digits the same? 12. Example. A nursery school teacher has nine children. a. how many ways can she line them up at the door to go for a drink? b. how may ways if little Lawrence had to be at the front or else he would cry? c. how many ways if Jackie and Dean are not allowed to be next to each other otherwise they fight? (Hint: calculate how many ways they can be together and subtract from the total unrestricted possibilities) Can you see why child care workers deserve a big hug!! LESSON 2 Permutations Involving Different Objects Objectives: To introduce the concept of a permutation as we have been developing. To find analytically the number of permutations of n distinct objects taken r at a time. Investigate

6 6 1. Two letters, A and B, can be written in two different ordered arrangements, AB and BA. These are called permutations of objects A and B. The number of ways to select things and arrange them in an order. a) List all of the permutations of 3 letters A, B, and C. How many permutations are there? b) List all of the permutations of 4 letters A, B, C, and D. How many permutations are there? c) Predict the number of permutations of 5 letters A, B, C, D, and E. 2. Instead of arranging letters in order, we can arrange objects if they are all different. a) How many different ways can 5 people be arranged in a line at the door for a fire drill? b) How many different ways can 6 different books be arranged on a shelf? What if we add the constraint that the dictionary had to be on the left though? c) How many permutations are there of the all the letters of the word PROVE? (1) What if we only wanted to make 4 letter permutations? (2) what if we only wanted to make 4 letter permutations that start with P 3. Consider the letters A, B, C, D, and E. Instead of using all the letters to form permutations, we could use fewer letters. For example, DB is a two-letter permutation of these 5 letters. a) List all the different two-letter permutations of the 5 letters A, B, C, D, and E. b) How many different three-letter permutations are there?

7 7 Example 1: 4. When you press the scramble or random button on an Mp3 player it plays a permutation of the songs on the Mp3. If the Mp3 has 5 songs on it, how many permutations of all the songs are possible (assuming of course the songs are all different!)? Example 2: 5. How many permutations can be formed using all of the letters of the word COMPUTE? Factorial Function Notation 6. Mathematicians who study this type of mathematics (called combinatorics or sometimes combinatronics ) got tired of showing expressions like all the time so they invented a special notation (a symbology) to write such an expression 2! = 3! = 4! = 5! = 20! = 49! = 69! = 70! = 0! = n! = Find the factorial button on your calculator!

8 8 7. What is the biggest factorial that your calculator can handle? (usually it is about 69 or 70) Properties of the Factorial Function : 8. The factorial function has some interesting properties: a. n! = n(n 1)! Eg: 6! = 6 5! = 6 ( ) b. n! = n(n 1)(n 2)! Eg: n! c. = n ( n 1)! 6! Example: = 6. Why? 5! d. The factorial function is only valid for positive integer numbers, despite that your calculator might give you an answer for numbers like 5.5! Example 3: 9. How many 3-letter permutations can be formed using the letters of the word COMPUTE? From the Fundamental Counting Principle we find that the number of permutations is: Remember, all we are doing is counting choices! Which we could also write it as So if we take the factorial of the total number of available objects; divided by the factorial of the total number of objects not chosen we can calculate the number of permutations of picking some things from a bunch of things. 11. The way a mathematician says this is: if n is the total number of objects and if r is the number of those objects that we want to pick out and arrange in some order then (n r) is the number of un-chosen objects. So the number of ways to make an ordered arrangement of r things out of a total of n things is: n! ( n r)! or 7! 4!

9 9 n! 12. Of course mathematicians have an even simpler notation for the expression ; they ( n r)! say n P r [ or some books or websites might use the notation P(n,r) ]. It doesn t matter how you indicate it but We pronounce n! n P r P( n, r) where n is the number of objects to pick from and r is the this number 7 permute of ( n r)! 3, or more often, those you want to pick. Obviously r must be less than n; since you can t pick out 7 more pick things 3. than there are total. 7! 7! Example: 7 P 3 = = = 765 = 210 (7 3)! 4! 4321 btw: the symbol above means: is defined as ; very much like equals but it is not making an equation, it is making a definition. We often tend to be sloppy and just use = a lot anyway. 14. Table of Permutations. A table of permutations is provided at the end of these class notes, sometimes seeing a table gives a better idea than just a formula. Check out the tables! 15. The Table of Permutations reveals that: a) permutations can get rather large and they do so very quickly b) the number of ways to select nothing (no things) from a bunch of things is one. ie: 5 P 0 = 1. Curious how selecting nothing counts as one possible arrangement. (So when my wife asked what I did this weekend from the family job jar, I can say I did one thing, because of the ten things she wanted me to do I did none, but there is one way to select nothing from the job jar!). Selecting nothing from any number of options is one thing in permutations! (don t try this argument at home; that nothing counts as something!) c) selecting an arrangement of n things from n things is the same number of ways as selecting (n 1) things from n things. ie: 10 P 10 is the same as 10 P 9. Can you explain why? 16. Summary: Permutations An ordered arrangement of objects is called a permutation. The number of permutations of all of n different objects is n! The number of permutations of n objects but chosen only r at a time is expressed by the formula: n! n P r = ( n r)!

10 10 Example: How many ways can you arrange 4 out of 6 books on a shelf if the order matters? 6! P 4 = = = 360 (Of course using the FCP you would have said: 6543 = 360) (6 4)! 2 Example 4: 17. Write each expression without using the factorial symbol to simplify it. n! a) n P 2 = = ( n 2)! ( n + 2)! b) = ( n 1)! Example 5: 18. Solve each equation for n. (these are rather disturbing but give them a try! Rather crazy algebra!) a) n P 2 = 42 b) 10 P n = 90 Definition of 0! 0! = Why?

11 11 LESSON 3 Permutations Involving Some Identical Objects (ie: Non-Distinguishable Objects ) Objective: To find the number of distinguishable arrangements of n objects when some of the objects are identical. Investigate 1. Exercise 1. The permutations of the 4 different letters A, B, E, and F are: ABEF ABFE AEBF AFBE AEFB AFEB BAEF BAFE EABF FABE EAFB FAEB BEAF BFAE EFAB FEAB EBAF FBAE BEFA BFEA EFBA FEBA EBFA FBEA How many arrangements are there? 2. a) What happens if two of the letters are the same? Investigate this by converting each F to an E in the list below. Then count the number of permutations of the letters A, B, E, and E. ABEF ABFE AEBF AFBE AEFB AFEB BAEF BAFE EABF FABE EAFB FAEB BEAF BFAE EFAB FEAB EBAF FBAE BEFA BFEA EFBA FEBA EBFA FBEA There are distinguishable arrangements of the letters A, B, E, and E. b) How does this number compare with Exercise 1? 3. a) What happens if three of the letters are the same? Investigate this by converting each F and E to a B. Then count the number of permutations of the letters A, B, B, and B. ABEF ABFE AEBF AFBE AEFB AFEB BAEF BAFE EABF FABE EAFB FAEB BEAF BFAE EFAB FEAB EBAF FBAE BEFA BFEA EFBA FEBA EBFA FBEA There are distinguishable permutations of the letters A, B, B, and B. b) How does this number compare with Exercise 1?

12 12 4. Generalize the pattern (make a general rule) from this investigation on the previous page to determine the number of distinguishable permutations of a) A, B, C, D, D b) A, B, D, D, D c) A, D, D, D, D d) A, B, B, C, C e) A, A, A, B, B Generalization The number of distinguishable permutations of n objects of which there are a objects alike of one kind, b alike of another kind, c alike of another kind, and so on, is: n! a! b! c! So what if you have 3 Blue Bingo Dabbers, 2 Green Bingo Dabbers, and 2 Red Bingo Dabbers? (all the exact same brand and shape so their colour is the only difference). How many distinguishable arrangements can you make of them in front of you in a line? a) Another way to look at this number of distinguishable permutations is to pretend that the Bingo dabbers have secret little numbers written on the bottom. So you have Blue 1, Blue 2, Blue 3, Green 1, Green 2, Red 1, and Red 2. b) If all seven Bingo dabbers were different colours you would have 7! ways to arrange them (5040 arrangements). Simple if they are actually all different. But since there are 3 blues which can be swapped around 3! ways (ie: 6 ways) then there will be only one-sixth of the arrangements would actually look different regardless of how you permuted the blues amongst themselves. Similarly, the 2 greens could be swapped about (permuted) 2! (or 2) ways amongst them selves, and so could the Reds.

13 13 Red 1 Green 1 Red 2 Blue 1 Green 2 Blue 2 Blue 3 c) above is one possible arrangement of the dabbers. For every way you can rearrange the bingo dabbers there are six ways that the blues look identical when the blues are permuted amongst themselves. So there will be six arrangements of blues that look identical for each and every arrangement of the all the bingo dabbers. So you need to divide by six (or 3! ) so you allow for the many ways that the arrangements look the same considering the blues. 7! d) The final result is that there is = 210 to distinguishably arrange the bingo 2!2!3! dabbers. Example 1: 6. Determine the number of (distinguishable) permutations of all the letters in each of the following words. a) MATHEMATICS b) OGOPOGO Example 2: 7. A true-false test has 7 questions and you are going to guess at all the questions since you did not study. But how many ways to answer are possible if someone tells you the test has 3 answers that are T and 4 answers are F? Does this help you much compared to if you did not know how many T and F there were? a) If you did not how many of each there were there would be ways to answer the test by completely guessing. b) Since you know there are 3 True and 4 False there would be arrangements. c) So does that improve your chances of guessing correct answers?

14 14 Example 3 - Pathways: 8. On the following grid (could be a city map), how many different paths can you take to get from A to B, assuming one can only travel east or south toward the general direction of B? Explain. A B Example 4. Pathways. 9. A mouse runs through a maze. It has six forks at which to make a decision, go left or right?! Investigate how many different paths the mouse can take! a. List them: (L, L, L, L, L, L), (L, L, L, L, L, R), (L, L, L, L, R, L),.[Do you want to keep listing them?] And how many of these paths does it end up at the same place? b. How many possible paths does the mouse have then? Use the FCP! Six selections to make, each has two choices! Answer: c. how many ways can the mouse select all lefts? d. how many ways can the mouse select all rights? e. how many ways can the mouse select three rights and three lefts?

15 15 f. how many ways can the mouse find the cheese? LESSON 4 Combinations Objectives: To introduce the concept of a combination. Combinations are generally more useful that permutations. To find the number of combinations of n distinct objects taken r at a time. Investigate 1. If five sprinters compete in a race, how many different ways can the medals for first, second and third place, be awarded? (hint: obviously the order matters!) Does the order of finish for the fastest three matter in how many arrangements are possible? This is an example of a permutation of objects taken at a time. 2. If 5 sprinters compete in a race and the fastest 3 qualify for the relay team at the Indigenous Games in New Zealand, how many different relay teams can be formed? Visualize the 5 sprinters below. Since 3 will qualify for the relay team and 2 will not, consider the number of distinguishable ways of arranging 3 Y s and 2 N s. Like three thumbs up and two thumbs down. Y Y Y N N Really, all you are doing is counting distinguishable arrangements of successes and failures. Like you had two different colours of Bingo dabbers

16 16 Does the order of finish for the fastest three matter here? This is an example of a combination of objects taken at a time. Combinations Summary An unordered arrangement of objects is called a combination. The number of combinations of n distinct objects taken r at a time is given by C r n n! n = although some references use other notation : C(n,r) or. ( n r)! r! r nc r will always be less than or equal to n P r. n Cr n Pr Examples 1: 3. How many different work groups of three people can be formed from seven people? Why is a combination lock not called a permutation lock? Doesn t the order of the numbers matter?? Just always wondered!!! 4. How many different committees of three people can be formed from seven people if one person selected serves as the chairperson, another as the treasurer, and the third as the secretary?

17 17 5. If the group of 7 people consists of 3 males and 4 females, how many different committees of 3 people can be formed with 1 male and 2 females? Think: you must choose 1 male out of the group of 3 males and 2 females out of the group of 4 females. 1 2 man women = 3 6 = 3 men 4 women 6. If the group of 7 people consists of 3 males and 4 females, how many different committees of 3 people can be formed with at least one male, that is: one or more males, on the committee? (hint to make easier: count what you are not looking for!) 7. If the teacher has enough money to take four students out of a class of ten to the burger joint on Taché and Marion: a. how many possible groups can he take? b. what if one of those four students had to be Norma (teacher s pet). 8. Properties of the Combination Function. A table of the combination function is attached at the end of the notes. Check it out, see what you notice! a) Ways to select nothing: (doing nothing counts as one thing in combinatorics! Every husband knows this!) b) Symmetry: c) Sum of any row (n).

18 18 Example Problems 2: 9. a) Evaluate(use a calculator also): 100! 3!97! b) Solve: nc 2 = 21 c) Evaluate 8 C 5 and compare it to 8 C 3. d) Evaluate 12 C 2 and compare to 12 C Lotto 6/49 consists of selecting a group of 6 numbered balls out of 49. Does order matter when the numbers are chosen? How many different lotto 6/49 combinations are there? Try this! Next time you buy a Lotto 6/49 ticket, ask the clerk for a Lotto 43/49! It is the same number of combinations! Whether you select the correct combination of balls that comes out of the lotto machine, or whether you select the remaining 43 that stay behind is the same thing! Think about it!

19 19 Example 3: 11. A standard deck of 52 playing cards consists of four suits (spades, hearts, diamonds, and clubs ) of 13 cards each. a) How many different five-card hands can be formed? n(five card hands)= b) How many different five-card hands can be formed that consist of all hearts (ie: a flush )? n(flush in hearts) = K Q J A Club Diamond Heart Spade c) How many different five-card hands can be formed that consist of all face cards? n (all face cards) = Can you make an outcome tree for these? d) How many different five-card hands can be formed that consist of three hearts and two spades? n(3h and 2S) = e) How many different five-card hands can be formed that consist of exactly four kings? n(4 Kings) =

20 20 f) How many different five-card hands can be formed that consist of at least four hearts? g) How many different five-card hands can be formed of at least one queen? Sometimes it is easier to count what you do not rather than what you do want! Especially if you only have two choices.

21 21 LESSON 5 Pascal s Triangle Objective: To develop Pascal s Triangle. Generation of Pascal s Triangle _ 0 _ 1 _ 1 _ 2 _ 1 _ 1 _ 0 _ 0 _ Observations 1. Relationship to n C r 2. The Symmetrical Pattern 5C 1 = Justification Generalization 3. The Recursive Pattern 5C 3 = Justification Generalization

22 22 Pathway Problems Revisited 1. On each grid, how many different paths can A take to get to B, assuming that you can only move to the right and down? a) b ) A How many blocks will you travel? How many will have to be right, how many down? A B B 2. If you have 10 points on a piece of paper and no three points are co-linear (meaning no three fall on the same line) then how many different triangles can you make? 3. How many ways are there to get from A to B (three dimensions now!) along the edges of the cube. B A 4. How many ways (paths) can you move down the rows of letters to spell PERMUTE. P E E R R R M M M M U U U T E T

23 23 LESSON 6 The Binomial Theorem Objectives: n To develop the Binomial Theorem expansion of ( a + b). To develop the general term formula for the (k + 1) th term in the expansion of n ( a + b). Investigate the Patterns in Binomial Powers 1. Expand and simplify each of the following powers of the binomial a + b using you knowledge of multiplying polynomials from Grade 9: (a + b) 2 = (a + b) 3 = (a + b) 4 = 2. What pattern do you see in the number of terms in the expansions above powers of a? powers of b? the numerical coefficients of the expansion terms?

24 24 3. Confirm your suspicion about the patterns by expanding and simplifying the following (done for you, but you may want to confirm that I did not mess up!): (a + b) 5 = (a+b)(a+b)(a+b)(a+b)(a+b) =(a+b)(a+b)(a+b) [(a 2 +2ab+b 2 )] =(a+b)(a+b) [a 3 +2a 2 b+ab 2 +a 2 b+2ab 2 +b 3 ] = (a+b)(a+b)[a 3 +3a 2 b+3ab 2 +b 3 ] =(a + b)(a 4 + 3a 3 b+3a 2 b 2 +ab 3 +a 3 b+3a 2 b 2 +3ab 3 +b 4 ) (a+b)(a 4 +4a 3 b + 6a 2 b 2 +4ab 3 +b 4 ) = a(a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 )+ b(a 4 + 3a 3 b+3a 2 b 2 +ab 3 +a 3 b+3a 2 b 2 +3ab 3 +b 4 ) =(a 5 +4a 4 b+6a 3 b 2 +4a 2 b 3 +ab 4 ) + (a 4 b+4a 3 b 2 +6a 2 b 3 +4ab 5 +b 5 ) =a 5 +5a 4 b+10a 3 b 2 +10a 2 b 3 +5ab 4 +b 5 Visualizing The Binomial Expansion of (a + b) 4 (a + b) 4 = 4 C 0 a 4 b C 1 a 3 b C 2 a 2 b C 3 a 1 b C 4 a 0 b 4 Example 1: 0 b s 1 b 2 b s 3 b s 4 b s Number of ways to choose this many b s from 4 factors of (a + b) 4 4. Expand by multiplying using the binomial expansion (you may want to check by making tables or graphs of your expansion): a) (x + 1) 4 = b = ( ) 3 =

25 25 b) (x + 3) 4 = c) (2x + 1) 4 = d) (2x 3) 4 =

26 26 e) 3 2 x + = x (to confirm: graph the two or make a table of values for different x to see if your expansion is the same) f) (0.95) 6 = (1 0.05) 6 = Pretty easy to check this with any calculator!(0.95 ^ 6 or 0.95 y x 6 depending on your calculator) The Binomial Theorem (using combinations) For any whole number n: (a + b) n = n C 0 a n b 0 + n C 1 a n-1 b 1 + n C 2 a n-2 b n C k a n-k b k + + nc n a 0 b n term k: 1 st term 2 nd term 3 rd term (k+1) th term (n+1) th term For this expansion the (k + 1) th term is: t k+1 =.

27 27 Example 2: 5. Determine the 4 th term in the expansion of (x 3) 9. Hint: the 4 th term is a k of 3. So 9 C 3 will be the primary coefficient in the expansion since 9C 0 is the first term. Example 3: 5. Write the first four terms of the binomial expansion of (x + 2y) 12.

28 28 Example 4: 6. The binomial expansion can sometimes be used to get very close answers to a problem if you have no calculator. For example, what if you invested $1000 for 8 years and compound interest was paid at a rate of 5% annually. This is the same as 1000( ) 8. So you can expand the power, but normally you only need to do the first couple terms, because after that the final terms (terms are things that add) just get so tiny to really count for much. You rapidly get down to such tiny fractions of a penny that you likely do not care! (BTW: fractions are kinda nice here as usual!!). Of course if you had a calculator it would be pretty easy to see how close you got! Example The volume of a cube is given by its edge cubed or edge 3. So if you increase the edges of a 10cm 10cm 10cm cube by 26% each edge then how much more volume does the cube hold (express as an exact number!). ie: find = (1 +.26) 3 =? (but of course, you do not have a calculator!) 100 Pretty easy to check if you used a calculator: = So a 26% increase in the edges pretty much doubles the volume. Notice just using the first two terms (a second order approximation) would tell you 1.78 times the volume and a third order approximation would be

29 29 Example 6 - Binomial Probability Distribution. Finding binomial terms becomes important in special probability problems where there are two possible outcomes and you want to know the probability of several of them happening. You know for example that the probability of any single car turning south on Arlington is 20% or 0.2. a. So what is the probability that all three cars in front of you are all turning onto Arlington. When you study probability you will learn the probability that all three cars turning is given by the 3 rd term of the expansion: 3C 0 (0.8) C 1 (0.8) 2 (0.2) + 3 C 2 (0.8)(0.2) C 3 (0.2) 3 so the probability of all three cars turning is 3 C 3 (0.2) 3 or or 0.8% or less than one in a hundred. Notice that the above expression adds up to one or 100%, so all possibilities of any number of cars turning are accounted for. b. what is the probability that if there are eight cars in front of you that exactly two and only two will turn left? Use pathways if necessary.

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31 1 SELECTED GLOSSARY FOR PERMUTATIONS AND COMBINATIONS Combination Factorial! Fundamental Counting Principle (FCP) outcome permutation The number of ways of selecting objects to form an unordered group is a combination. Example: how many ways can 3 players of a 10 person team be selected for a league all-star game? It doesn t matter what order you get selected in, just that you make the team. The answer is 10 C 3 or 120 ways Compare to permutation. A special notation to show that a succession of descending integers down to one are multiplied together. n! n ( n 1) ( n 2) ( n 3)...1 eg: 5! = = 120 if one item can be selected in m different ways and a second item can be selected in n different ways, then the two items can be selected in m n different ways. Eg: if I have a choice of 3 pairs of pants and 6 shirts, there are 36 = 18 different outfits I can wear. a possible result of an experiment; a possible answer to a survey question. For the experiment of tossing a six-sided die, the possible outcomes are rolling a 1, 2, 3, 4, 5, or 6. Or selecting from a choice of options: should I wear my blue shoes or my brown shoes? An ordered arrangement of different objects. If all quantity n of some different objects are arranged there is n! ( n factorial ) ways to arrange them. If only a sub-quantity r ; where r<n, of the possible n objects are selected then there are n P r ways to arrange the objects where: n! n P r ( n r)! Examples: a. Five different bingo dabbers; arrange all five of them in front of you in a line. The number of ways is 5! or n! or 5 P5 = 120 (5 5)! b. Five different bingo dabbers, but you are only going to put three of them in front of you. How many arrangements? 5! or 5 P3 = = 60 (5 3)! 21

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33 Annex 1 A Table of the Permutation Function n P r [ or P(n, r) in some text books] npr r n ,520 5,040 5, ,680 6,720 20,160 40,320 40, ,024 15,120 60, , , , ,040 30, , ,800 1,814,400 3,628,800 3,628,800 Example Usage: How many ways can you seat 5 people, r, from a total group size, n, of all 5 people in an ordered arrangement on a bench. 5! Answer: 5 P 5 = = = 120 0! 1 How many ways can 10 runners take 1 st, 2 nd, and 3 rd place in a race. 10 P 3 = 720 (in an ordered arrangement: 1 st, 2 nd, and 3 rd ) Numbers can get rather big very quickly with permutations. The ARRANGEMENT ORDER MATTERS with Permutations. Patterns. Seeing as math is the study of patterns what other patterns do you notice?

34 Annex 2 Combinations Formula Table. n C r ncr r n ,287 1,716 1,716 1, ,001 2,002 3,003 3,432 3,003 2,002 1, ,365 3,003 5,005 6,435 6,435 5,005 3,003 1, Patterns: a. Symmetric in each row. b. The sum of each row is 2 n. c. nc r = n-1 C r-1 + n-1 C r. d. And even more pattern if you study someone called Fibonacci!! Example Usage: a. Combinations. How many ways can a committee (an unordered group) of 4 be selected from 12 people? n = 12, r = 4. So: 12C 4 = 495 arrangements where order does not matter. b. Pathways. Do you notice how this table relates to pathway problems?? If you have 10 steps to make, and 4 of them have to be left and 6 down, then there are 10 C 6 ways to select a combination of steps. ORDER DOESN T MATTER with Combinations. Like the Lotto 6/49! Symmetry. Notice how 6 C 2 is the same as 6 C 4. Notice how 10 C 1 is the same a 10 C 9. Notice how 3 C 2 is the same as 3 C 1. Notice how 9C 0 is the same as 9 C 9. And the number of ways you can choose do select nothing is the same as the number of ways you can choose to select all things. See a pattern? n C r is the same as n C (n-r)