Lecture 31 21 127 Concepts of Math 11 14 2012 Counting In Two Ways Method Outline We are given an equality and we must explain why it holds True To do this we have to define a set whose size is represented on both sides of the equation and then explain why it all works out Usually the tricky part of these problems is figuring out what that set is in the first place We ll do several examples today and try to show how we approach coming up with an interpretation for the two sides of the equality In general we re looking to tell a story and describe multiple processes and or partitions that produce the same set of outcomes For example you did these in recitation Pascal s Identity n n n 1 k k 1 k 1 Proof summary Partition the number of ways to invite k 1 people out of n 1 total friends to a party based on whether or not I invite my particular friend Joe Alternatively count the number of subsets of n 1 with size k 1 based on whether or not the subset has 1 as an element Alternatively count the number of binary strings of length n 1 that have exactly k 1 0s by partitioning based on whether or not the first position is a 0 This identity is why Pascal s Triangle works Chairperson Identity n n 1 k n k k 1 Proof summary Choose a committee of size k from n total people with a designated president do this by selecting the committee members first and then making one of them the president or do this by selecting the president from the entire pool and then filling in the rest of the committee Alternatively count the number of binary strings of length n with exactly k 0s and with one of those 0s circled do this directly or by placing the circled 0 first and then placing the rest of the 0s Alternatively count the number of subsets of n with size k that have one element circled do this directly or by selecting the circled element first and then selecting the rest Method Outline 1 State the result to be proven Note remember to quantify any variables that appear in the expression 2 Define a set let s call it S of objects to be counted 3 Count the elements of S in one way by following a proper combinatorial argument Equate the derived expression with S 4 Count the elements of S in another way by following a proper combinatorial argument Equate the derived expression with S 1 5 Conclude that since both derived expressions equal S they must be equal as well Common mistakes to avoid Forgetting to define a set of objects to be counted Defining a set but counting something else in two ways Counting one set of objects but then counting a different set of objects in another way Failing to equate the two expressions in a conclusion Examples The Two Class Schedule Identity Say there are m different math classes and p different physics classes to choose from Assume there is no cross listing I need to take exactly 2 classes to be a registered student and keep getting a paycheck How many schedules can I create m p m m p p 2 2 1 1 2 Intuition From the pool of all m p classes pick two Or partition based on whether I take two classes from one department or one from each Proof Let S be the set of possible 2 class schedules I can create Since the total number of courses to choose from is m p and I am selecting 2 S m p 2 Consider the partition S A B C where A is the set of all schedules with 2 math classes B is the set of all schedules with 1 course from each department and C is the set of all schedules with 2 physics classes Certainly this is a partition of S since every schedule satisfies exactly one of these m m p p criteria By definition of selection and ROP we see that A and B 2 1 1 and C 2 m p p Then by ROS we see that S m 2 1 1 2 Since we found S in two ways we deduce that the claimed equality holds Pro Con Committee Identity n i n n k X n 2 i k k i k Intuition Create a committee of size k from n people Then determine whether the non committee people are for or against the committee s decisions We could also do this by first selecting at least k people who will be on for the committee and set everyone else to be off and against Then from that pool we select k people to actually be on the committee setting the others to be for it Note It s important to say all of the steps we will perform in these constructions Don t assume anything is obvious to the reader Proof Consider a set of n people Let S be the set of ways to select k of the n people to be on a committee with every person who isn t on the committee having a firm opinion to be For or Against the committee First we can find S by a multi step process Select k of the n people to be on the committee n k options 2 For each of the remaining n k people have them decide whether to be For or Against This is a process with n k steps and two choices at each step so by ROP 2n k options By ROP we have S n k 2n k Second we can find S by establishing a partition based on how many people are For the committee By the definition of S anywhere from none to all of the n k non committee members could be For the committee In total then between the committee members and their For supporters we can have somewhere from k to k n k n people inclusive For each i that satisfies k i n let Si S be the set of ways to have k committee members and i k For supporters Notice that 0 i k n k which matches the restriction noted previously Notice that Si k i n is a partition of S This is because for any element of S that element can be characterized by how many For supporters of the committee there are and that has to be some specific number Now we can find Si for each such i by a multi step process From all n people select i people These are potential committee candidates n i options Designate the n i other people to be decidedly Against the committee we are constructing This step is deterministic so there is only 1 way to do it but we need to point this out to fully describe an outcome that is an element of S Of those i people …