Lecture 30 21 127 Concepts of Math 11 12 2012 Administrivia Homework 9 is still due Thursday Remember that you cannot do any dividing out or subtracting out for overcounting We are stressing ROS and ROP so that we learn why those methods work Homework 8 will be returned tomorrow Here are some comments I have about the problem I am grading Be careful with the word only in logical claims This means that a certain claim holds true in that situation and no others Many of you wrote A union of countably infinite sets is only countably infinite if the number of sets in the union is countably infinite or finite Or something like that This is flat out False You even showed why on Problem 2e and I even made a comment on the AnnotateMyPDF site on this problem to say Perhaps you want to solve Problem 2e first and think about that here That is a verbatim quote by the way For instance N N x R is an uncountable union of infinite sets that is countable The logical issue is like this Any prime p 3 is odd This is a True claim What does it have to say about this situation Notice p 15 and p 3 and p is not prime Therefore it is even Bwaaa Of course not That s silly But that s exactly the same logic many of you employed just in different terms When we prove P Q this does not necessarily mean Q P too Knowing a certain situation yields a conclusion does not necessarily mean that no other situations yield the same conclusion Counting Problems Example How many 4 tuples from 0 1 2 have at least one of each number Standard overcounts Pick a spot for the 0 pick a spot for the 1 pick a spot for the 2 then pick one of the three to appear in the empty spot 4 3 2 3 1 1 1 1 Pick 3 spots to be filled with 0 1 2 Permute them in those spots Pick a number for the 4th 4 3 3 3 We can exhibit specific examples that are counted in multiple ways to show that these counts are wrong For the second example consider the 4 tuple 2 0 1 0 This could have arisen from the process specified in at least two ways 1 We could have selected the first three spots in Step 1 Then we could have chosen to fill those spots with 2 and 0 and 1 in that order Then we could have filled the remaining spot the fourth with a 0 Alternatively we could have selected the first third and fourth spots in Step 1 Then we could have chosen to fill them with 2 and 1 and 0 in that order Then we could have chosen to fill the remaining spot the second with a 0 This shows that the same outcome is considered by this process more than once so it is an overcount We won t even think about how to correct this bad count In the context of this example we don t even care the point is to show how we can find flaws in counting arguments by exhibiting particular outcomes Example Consider rolling 2 dice How many outcomes are there It depends on how we interpret this and how the information is presented If we think of this as rolling one die and then rolling it again there are 6 6 36 outcomes This is like representing the outcome as an ordered pair of numbers each of which is from the set 1 2 3 4 5 6 For example if we rolled a 5 and then another 5 we would have the outcome 5 5 If we think of this as someone presenting us with two unordered numbers as the outcome of some experiment then there are fewer than 36 outcomes If we obtain the information in this way we cannot distinguish between the dice All we re getting here is an unordered pair of numbers Alternatively we can think of this as a multi set where order does not matter but repeats are allowed For example if the experimenter rolled a 5 and a 5 he would just say There is a 5 and a 5 Notice he did not say and then a 5 like we did above there is no inherent distinguishability of the dice so we can t tell them apart In this second interpretation we can count the possible outcomes by partitioning based on whether a number repeats or not and applying ROS 6 6 6 6 5 6 6 15 6 21 2 1 2 4 2 The first term is the number of outcomes with two different numbers and the second term is the number of outcomes with just one number Notice that this process treats the ordered pairs 4 6 and 6 4 as the same whereas the first interpretation treated them differently Remember that we don t know how to divide out or subtract for overcounting We only have the ROS and ROP at our disposal because we proved them The next two examples emphasize this distinction Example Consider rolling 4 distinguishable dice so we can assume they re colored say Red and Yellow and Green and Blue What does the following formula represent 64 0 4 2 2 4 0 4 3 3 4 3 3 4 3 3 0 1 1 2 1 1 4 1 1 This seems hard It would be better to present this as a sum and deduce a partition Let s say the above number is S where S is some set of outcomes Then 0 4 2 2 4 0 4 3 3 4 3 3 4 3 3 S 64 0 1 1 2 1 1 4 1 1 It looks like we are partitioning the set of all outcomes of rolling the 4 distinguishable dice into 4 cases In the three terms with binomial coefficients it looks like we choose 0 or 2 or 4 of the dice and then do 2 something The exponents on the 31 terms always sum to 4 and one of them is always the bottom part of the k4 term Thus it looks like we are selecting some of the dice assigning the numbers on them somehow and then assigning the numbers on the other dice If you think about it for a minute these three terms represent the number of ways we could have an even number of dice showing an even number on their face after rolling them We select 0 or 2 or 4 of the dice then select one of the 3 even numbers 2 or 4 or 6 they could show For the 4 k dice left over we select one of the 3 odd numbers 1 or 3 or 5 they could show These sets of outcomes together with the set S of outcomes where there are an odd number of even numbers showing partition the set of all outcomes Answer The above formula represents the number of outcomes …