Exam 2 Wednesday, October 19, 2016 11:53 AMContents: - Counting and tree diagrams - Rearrangements - probability Counting and tree diagrams: Suppose that you have the set S= {1,2,3,4} How many 3 element subsets can one create? A: 4 Use a tree diagram Levels represent a selection round Leaves represent an outcome

Since we don’t care about the order you don't count all the leaves (the definition of a subset is that the elements are the same no matter the order. Ex. 1234 and 2314) With bigger alphabets you can figure out the answer by taking the number of total rearrangements and dividing it by the number of nce we on care aou e orer you on coun a the leaves (the definition of a subset is that the elements are the same no matter the order. Ex. 1234 and 2314) With bigger alphabets you can figure out the answer by taking the number of total rearrangements and dividing it by the number of subsets. Ex. If you have the alphabet {0,1,2,3,4,5,6,7,8,9} and you want to find the number of 4 element subsets that can be created with that set: First: there are 10x9x8x7 possible 4 element arrangements Then figure out that there are 24 possible subsets because 1 subset contains 4 elements that can be arranged 4! Ways Now let's try a harder example:How many codes of length 7 can you create using the alphabet {0,1,2,3,4,5,6,7,8,9} (repeated digits allowed)
First Step:
Determine the size of your alphabet You have a 10 digit alphabet

Step two: Determine how many levels of the tree you will have You have 7 choices for each digit in the code. Raise 10 to the 7th because you have 5 choices

## How many ways are there to rearrange the word CHOICE?

## How many 5 digit codes can you create with {0,1,2,3,4,5,6,7,8,9} with no consecutive digits?

## How many 3 element subsets can one create?

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Let's make it even harder: Restriction is that repeating digits are not allowed

Step one:

Step two: As in example 1 determine the size of your alphabet and how many choices you have Since you have 7 choices it may help to create a visual aid: _ _ _ _ _ _ _ _ In each of the dashes you have one digit two:create a visual aid: _ _ _ _ _ _ _ _ In each of the dashes you have one digit of the code. Underneath you can place how many choices you have for that code For the first one you have 10 choices because you haven't used any For the second one you have 9 choices because you have repeated one already Step 3: When you discover how many choices you have for all the digits use the Rule of Multiplication to find the answer (10x9x8x7x6x5x4) Now you try: How many 5 digit codes can you create with {0,1,2,3,4,5,6,7,8,9} with no consecutive digits? Rearrangements:Here we take into account the order of the digits Note: factorial notation (!) is simply multiplying a digit until you reach 1 (i.e. 4x3x2x1) How many ways are there to rearrange the word CHOICE? Since none of the letters repeat we do the same as in the code example You have 5 letter placements. The first spot has 5 letter choices, the second 4, and so on. So the answer would be 5! Or 5x4x3x2x1 How many ways can you rearrange the word LENTILS? First you must find how many ways you can rearrange the word if all letters are distinguished. The word has 7 letters so there are 7! Ways of rearranging them. Then divide the answer with the possible rs you mus n ow many ways you can rearrange the word if all letters are distinguished. The word has 7 letters so there are 7! Ways of rearranging them. Then divide the answer with the possible ways of rearranging the repeating letters (here it would be the L's) So the answer would be !!#!⎯⎯ How many ways can you rearrange WISCONSIN? First you have 9 letters so you can rearrange them 9! Ways But, you have 3 pairs of recurring letters (S,I,N) so you can rearrange those 2!x2!x2! Ways The answer is 9!/ 8 How many ways can you rearrange UNUSUAL? 7! /3! (3 U's) How many rearrangements have the 3 U's together? 5x 4! (5 choices to arrange Us and 4! For the other letters) How many arrangements have no consecutive Us? 5x4x3 (places for Us) / 3! (ways to arrange the Us) x 4! (ways to arrange remaining letters) Now you try (answer key at the end): How many ways can you rearrange NOTATION? How many ways can you rearrange HAWAII? How many ways can you arrange MISSISSIPPI? How many 8 digit numbers can you make with the following [3,4,5,6,7} ? 5& How many are larger than 50000000? 3x 5! (Because for the first choice you only have 3 numbers that are equal or greater than 5) Suppose you have 5 keys on a table and you pick 3x (Because for the first choice you only have 3 numbers that are equal or greater than 5) Suppose you have 5 keys on a table and you pick them up one at a time and place them on a key ring. How many different configurations are there? 5/5! Because in a circular rearrangement you only care about relative placement (neighbors) Probability: Suppose you have a "fair" coin and label the sides H and T. Suppose you flip this coin 5 times and keep a record of what happens. Ex. HTHHT represents flips 1-5 How many sequences are there? 2( In how many of the sequences have the property where 4 H's and 1 T occur in the sequence? To know the answer you find how many ways to arrange the T. So the answer is 5 Suppose you write down all possible sequences of 4 digit codes created by the alphabet {0,1,2,3,4,5,6,7,8,9} (repeated letters allowed) then put them in a hat and mix them up. How many slips of paper all there? 10+ What is the probability that we will pick a paper that has 34XY (X,Y could be anything)? 1/100 because you divide the number of elements in the event (in this case 34XY) by the total number of elements in the set (10000) Why is Pr[E] + Pr[E'] = 1? Because an event can either happen or not (1 means very likely to happen) In this formula it states that the event plus it's compliment (not the event) are very likely to happen Let E be the event: You draw two marbles from a jar at random that contains 3 marbles (1 blue, 1 pink, and 1 orane and don't relace the first marble before ickin compliment (not the event) are very likely to happen Let E be the event: You draw two marbles from a jar at random that contains 3 marbles (1 blue, 1 pink, and 1 orange) and don't replace the first marble before picking the second. What is the probability you will draw a pink and blue marble? 1/3(probability of the first ball) x 1/2probability of the second ball) = 1/6 Now suppose the jar contains 4 blue marbles, 3 pink marbles and 3 orange marbles. What is the probability of you picking a pink and blue marble? 3/10 x 4/9 = 12/90 Answer Key:5 digit codes: 10x9x9x9x9 NOTATION: 8! / 2!x2!x2! HAWAII: 6! / 2!x2! MISSISIPI: 11!/ 4!x4!x2!