Chapter 10A. a) How many labels for Product A are required? Solution: ABC ACB BCA BAC CAB CBA. There are 6 different possible labels.

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1 Chapter 10A The Addition rule: If there are n ways of performing operation A and m ways of performing operation B, then there are n + m ways of performing A or B. Note: In this case or means to add. Eg. You have to order a set of simple identification labels for two different kinds of products. Product A requires an id label that is made up of 3 letters (from A to C), and Product B needs labels of 2 numbers (from 1 to 5). (no doubling up of letters or numbers is permitted) a) How many labels for Product A are required? Solution: ABC ACB BCA BAC CAB CBA There are 6 different possible labels. b) How many labels for Product B are required? Solution: There are 20 different possible labels c) How many different ID labels need to be ordered for all possible arrangements. Solution: Total labels = Possible Product A labels + possible Product B labels = = 26 You have to order 26 different labels.

2 Task 1 You are going to the school dance. You only have 2 pairs of dancing shoes, one black pair and one brown pair. To go with your black shoes you have 3 shirts and 2 pants, while the only things that match your brown shoes are 2 shirts and 1 pair of pants. a) If you decide to wear your black shoes, what are the total number of combinations of clothes that you can wear? (draw a tree diagram) b) If you decide to wear your brown pants, what are the total number of combinations of clothes that you can wear? (draw a tree diagram) Ans: 6 c) What are the TOTAL number of possible outfit options you have? Ans: 2 Ans: 8

3 The Multiplication rule: If there are n ways of performing operation A and m ways of performing operation B, then there are n m ways of performing A and B. Note: In this case and means to multiply. Eg. I have 6 shirts and 3 pants. What are the total number of combinations? Solution: 6 x 3 = 18 You can verify this by drawing a tree diagram.

4 Task 2 Sam needs to go over to Tim s house, but on the way he needs to pick up Cam and take him also. There are 3 different ways to get from Sam s house to Cam s, and then 4 different ways to get from Cam s over to Tim s. a) How many different ways could Sam go. Task 3 Sam, Cam and Tim are eating out and there is a selection of 2 entrees, 3 mains, and 3 desserts. a) Sam only has entrée and dessert. How many total possible different combinations of meals could he have? Ans: 12 b) Tim has a main and a dessert, how many total possible different combinations of meals can he have? Ans: 6 c) Cam has all three. How many total possible different combination of meals can he have? Ans: 9 Ans: 18

5 We now need to extend the multiplication rule a bit further Task 4 You have 2 different coins. How many different ways can you line them up? How many options do you have for the first coin Now, how many options do you have for the second coin Using the Multiplication Rule, How many arrangements can you have? You now have 3 different coins. How many ways can you arrange them? # options for 1 st coin # options for 2 nd coin # options for 3 rd coin Multiplication Rule gets our answer You now have 4 different coins. How many ways can you arrange them? 1 st coin options 2 nd coin options 3 rd coin options 4 th coin options Use the rule to get What is the pattern here? What is your hypothesis about the ways of arranging n different items? Now, lets put this into Practice and attack Chapter 10A with zeal and gusto. Humans learn by doing. Practice is Everything, so why not just do them all J

6 Chapter 10B Lets talk Factorials. The easiest way is to just look at an example. 5! = We would say this is Five Factorial. So in essence it is an abbreviation of a recurring and reducing multiplication. Mathematically we could show this as n!= n ( n 1) ( n 2) ( n 3) Possibly this looks much like your hypothesis of arranging n different things from the last lesson? Factorials are deceptively concise. The abbreviated way they are written defy how quickly their value increases. Mentally calculate Three Factorial 3!= =? How about Five Factorial 5!= =? Surely Eight Factorial can t be too big? 8! = =? Lets make sure we know how to calculate Factorials in your Calculator!!! nb: 0! = 1 yes, this seems weird, but I have proof that 0! = 1, and here it is: Since n! = n(n 1)(n 2)(n 3) so we could also say n! = n n 1! Setting n = 1, we get 1! = 1 1 1! 1 = 1 0! 1 = 0! So clearly, 0! = 1 Just because I know you love it, take a look at Now, where would we use this?

7 Task 5 Think about a horse race and all the possible ways they could finish. a) If there are three horses, what is the total number of possible ways they could come across the finish line? 1 st place possibilities 2 nd place possibilities 3 rd place possibilities Apply the multiplication rule b) if there are 5 horses? 1 st 2 nd 3 rd 4 th 5 th Multiplication Rule c) If there are 9 horses? (take the short cut and use the factorial computation) d) What about 15 horses? (The odds are stacked against the punter!) Can you see the pattern? We can now establish another Rule! ** The number of ways that n distinct objects may be arranged is n! If only it were this simple, but we do not always simply arrange everything we have.

8 Task 6 Lets return to the idea of arranging our coins. Assume we have one of each coin, that is; $2, $1, 50c, 20c, 10c and 5c. a) How many ways can you arrange 2 of these coins in a line? # 1 st coin options # 2 nd coin options Apply Multiplication rule b) How many ways can you arrange a series of any 3 of these coins? 1 st coin options 2 nd coin options 3 rd coin options Apply Multiplication rule c) How many ways can you arrange a series of any 4 of these coins? 1 st coin 2 nd coin 3 rd coin 4 th coin # Arrangements As you can see, these are Not Full Factorials, but stop at some point through the factorial process. As the relationship between the number of objects and how many in the arrangement, we can use a Rule. The Permutation Rule: The number of permutations (different arrangements) when r things are chosen from n things and order is important is:!p! =! P! = P!! = n! n r! (You will see! P! written many different ways depending on which text book or web page you are looking at. They are all the same!) Return to the above tasks, and re-do them using the rule and double check your calculations. In your Ti-nSpire menu, Or simply type npr(, ) straight in

9 Circular Permutations: Sitting in a circle has its own issues. There is a special formula when working permutations in a circle.!p! r Now, lets put this into Practice. Exercise 10B. Humans learn by doing. Practice is Everything, so why not just do them all J

10 Chapter 10C Some adjustments need to be applied when restrictions are added to arrangement requirements. Some restrictions need strategic adjustments to the determination of n and r ; Eg. Ten students are in a race. Bill is State champion so he is a sure winner, but the others are very close. Calculate the total number of possible podium finish placings. Solution: We need to withdraw Bill from the group and place him in first position. This will effect our value for n which will now NOT be 10, as well as our value for r which will now not be 3. Here, n = 9 and r = 2 things using the formula P r n = n! n r ( )! by substitution we get P 2 9 = 9! ( 9 2)! = 9! 7! = 72

11 There are also possible indistinguishable arrangements. Where this happens we will use a Grouping technique. This is easiest to describe with an example: One use of the grouping rule is with letters. Clearly if you have duplicate letters, then they can be transposed and make no difference to the arrangement, such as the letters A, B, A, B. Here we can label them A 1 B 1 A 2 B 2 Clearly the following arrangements are the same : A 1 B 1 B 2 A 2 A 2 B 1 B 2 A 1 A 1 B 2 B 1 A 2 etc So we need to make adjustments to our solutions. The Grouping Rule: The number of different ways of arranging n things made up of groups of indistinguishable things, n 1 in the first group, n 2 in the second group and so on, is; = n! n 1!n 2!n 3!...n r! Eg. How many different arrangements can be made using the 11 letters of the word Mississippi? Solution: No. of letters n =11 No. of I s n 1 = 4, No. of S s n 2 = 4, No. of P s n 3 = 2 TotalArrangements = n! n 1!n 2!n 3! = 11! 4!4!2! = 34,650 Now, lets put this into Practice. Exercise 10C. Humans learn by doing. Practice is Everything, so why not just do them all J

12 Chapter 10D - Combinations The difference between Permutations and Combinations, is that in one the order DOES matter, whereas in the other, order does NOT matter. Permutation: Selecting a Chairperson, Secretary and Treasurer of a committee from 10 applicants Order Matters Combination: Selecting a team of three delegates from 10 applicants order does Not matter Other examples would be dealing cards, as the order in which you receive the cards, makes no difference to the hand that you end up with. Because order does NOT matter, we need to cancel out the duplicated entries that the Permutation Formula gives us. Refer to page 446 for how this is derived.!c! = C!! = n r =! P! r! = n! n r! r! ( again, you will see ncr written many different ways. I have no idea who came up with the weird bracket thing J ) In your Ti-nSpire menu, 5., 3. or simply type ncr(, ) Now, lets put this into Practice. Exercise 10D. Humans learn by doing. Practice is Everything, so why not just do them all (even come back and do this extra one J ) There are 15 students and they have to get into 3 evenly numbered groups. In how many ways can the groups be formed?

13 Firstly, as we are putting people in General groups, then order does Not matter, so we are clearly talking about Combinations! There will be a certain number of ways to form the first group out of the whole group, then a certain number of ways to form the next group out who is left, and then the last group. Then, like doing a probability tree-diagram, we can apply the multiplication rule J Group 1, is selecting 5 students out of 15!"C! Group 2, is selecting 5 students out of the remaining 1!"C! Group 3, is selecting 5 students out of the remaining 5!C! Then applying the Multiplication rule: Total Combinations =!" C!!" C!! C! So the total combinations is = 756,756

14 Summary Factorials n! = n n 1 n 2 n NOTE: 0! = 1 Permutations - With permutations, order DOES matter. Permutation of n objects is simply n! Permutations of r objects from n!p! = n! n r! Circular permutations If r people out of n are to be seated in a circle:!!!! Permutations with duplications n! n 1!n 2!n 3!...n r! eg. How many arrangements can you make from the letters in Parallel? Noting there are 8 letters, of which there are 2 A s and 3 L s, we get: n = 8 n 1 = 2 n 2 = 3 solution 8! 2!3! = 3360 Combinations - With combinations, order does NOT matter. Combinations of r objects from n!c! = n r = n! n r! r! Lets put these into practice Exercises 10E yep, you guessed it All of em!

15 Chapter 10F The Binomial theorem Pigeon hole yuk lets just look at the Binomial Theorem! *** We are doing the BiNomial Distribution in Maths B at the moment. This Binomial is Completely Different. Do NOT get them mixed up!!! *** Here, binomial is a special type of polynomial. Bi is the prefix for Two, so a Binomial has Two terms, such as x+3. I trust you are aware of Pascal s triangle, if not, then take a look at page 463? ** nb: the top row, is the 0 th row!!! The numbers in the rows, correspond to the coefficients in an expansion to the power of the row. So Row 2 is raising a + b! Pascals Row 2 is Hence, the expansion becomes 1a! + 2ab + 1b! Clearly, if either a or b are values, or have coefficients, then the simple use of Pascals triangle becomes difficult. Luckily we have a rule for that! a + b! = a! +! C! a!!! b! + b! So, in the expansion of x + 7!, the 4 th term would be!c! x!!! 7! = 56 x! 343 = 19208x! Seems simple enough??? So, last chapter get into some Exercise 10F questions You need to know this content Thoroughly, so you can apply the correct rule and the correct process in the right situation. You need to be able to APPLY your understanding in unfamiliar situations. This can only be done if your knowledge is deep and thorough!

16 Some Revision I have 6 different coins. $2, $1, $0.50, $0.20, $0.10 and $ How many different ways can I arrange them? 2. I only have room to display 4 coins. How many ways can I arrange 4 of them? 3. I want to select three coins. How many different overall values of money can I obtain? 4. I have 3 kids. I want to give one child, one coin, a different child two coins and the last child three coins. How many ways can I share my $3.85? 5. How many distinct arrangements of the letters on the word HORROR can I have? Answers:

17 Now to something more Abstract : Evaluate:!!!P!!!!C! Ans: x(x + 1) and! x(x + 1)! Before you do the next question, given the difference to the previous question, test your understanding to consider what you think will be the difference in the outcome?!!!p!!!!c! Or, how about something like Ans: (x 2)(x 1) and! (x 2)(x 1)!!!!P!!! or even!!!p!!!!c! Ans:! and!!(!!!)!!!!!! (!!!)(!!!) or how about an actual equation, like Find n, such that: 3! C! = 30 Answer, n = 5, or 4

18 A Maths B class has 10 boys and 12 girls. I need a committee of 3 students to organise an engineering trip. a) How many possible committees can I have? b) What if I only want to pick girls. (as they are more reliable?) c) OK, that s not appropriate! How about I have 2 girls and 1 boy? Ans: 1540, 220, 660 I have six colours; red, white, blue, green, yellow and black. Find the number of two colour combinations. Ans: 15 Have you ever played Bridge or 500? Each player gets 13 cards. Playing cards is pretty basic isn t it J Lets deal a deck of cards How many different possible hands can I have? Ans: !! Shuffle an Imaginary deck of cards. Your have just done a worlds first. No other deck of cards has Ever been in that particular order J You think I am joking don t you? Make sure you watch this! and/or