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10x9x8x7x6x5x4

10x9x8x7x6x5x4

Description

School: American University
Department: Applied Mathematics
Course: Finite Mathematics
Professor: Chris mitchell
Term: Fall 2016
Tags: Math Finite, Probability, rearrangements, and treediagrams
Cost: 50
Name: Finite 150 Exam 2 Study Guide
Description: These are examples of the material that is on the exam. Some are re-worked examples we got in class
Uploaded: 10/24/2016
6 Pages 160 Views 1 Unlocks
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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!
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