Lecture 33 21 127 Concepts of Math 11 19 2012 Administrivia Exam 3 Prep Questions are posted Exam 3 is on Friday 11 30 Please check your homework grades to make sure they are correct Send me an email about any errors I also posted a short story by Jorge Luis Borges on AnnotateMyPDF under Extra Readings It s entitled The Library Of Babel Read it It s very interesting And related to what we re doing now Each wall of each hexagon is furnished with five bookshelves each bookshelf holds thirty two books identical in format each book contains four hundred ten pages each page forty lines each line approximately 80 black letters There are also letters on the front cover of each book those letters neither indicate nor prefigure what the pages inside will say There are twenty five orthographic symbols Footnote The original manuscript has neither numbers nor capital letters punctuation is limited to the comma and the period Those two marks the space and the twenty two letters of the alphabet are the twenty five sufficient symbols that our unknown author is referring to Counting Objects Lattice Paths Another type of counting object to be used Lattice paths Let x y N 0 2 A lattice path to x y is an ordered tuple of points in the plane lattice where the first element of the tuple is 0 0 the last element of the tuple is a b and every element in the tuple only differs from the previous one by having exactly one coordinate that is exactly one larger than the corresponding coordinate of the previous element More rigorously given x y a lattice path is an n tuple P1 P2 Pn for some n N where each Pi xi yi is a point in the lattice and i n 1 xi 1 yi 1 xi 1 yi xi 1 yi 1 xi yi 1 and furthermore x1 y1 0 0 and xn yn x y That is a lattice path is a sequence of points in the lattice from 0 0 to x y where we are only allowed to move rightwards or upwards by one grid point at every step To describe a lattice path from 0 0 to x y we know we will have a sequence of total length x y with exactly x rights and exactly y ups Thus there are x y x y x y total lattice paths This is exactly counting the number of binary strings of length x y with exactly x 0s where we interpret a 0 as move right and a 1 as move up 1 Example Let s look at the lattice paths from 0 0 to 4 2 How many are there 4 2 6 15 2 4 How many go through the point 1 1 Think of a two step process 1 Make a path from 0 0 to 1 1 2 Make a path from 1 1 to 4 2 By ROP there are 1 1 3 1 2 4 8 1 1 How many don t go through that point 15 8 7 Interestingly we can use Lattice Paths to prove the Summation Identity in a different way Proof Count Lattice Paths from 0 0 to k 1 n k by partitioning based on where the first Right step happens Let P be the set of all such paths For 0 j n k let Pj be the set of paths that first take a Right step at height j i e the paths that go up to 0 j and then to 1 j and then to the terminal point somehow Notice that k n k j n j Sj k k because we are at 1 j and need to go k more to the right and n k j more up Thus S n k X Sj j 0 n k X j 0 n j k Now we reindex the sum by setting i n j Then 0 j n k 0 n i n k 0 i n k n n i k Thus S n X i i k k n tuples Definition Let n k N be given We define Tn k to be the set of all n tuples drawn from the set k 1 2 k That is Tn k is the set of all ordered lists of length n each of whose coordinates is an element of k We can also write Tn k k n Example Consider the elements of T4 3 How many are there 34 81 How many have their coordinates sum to an even number To ensure this we need either 0 or 2 or 4 odd numbers If we have an odd number of odds then the sum is odd The only even number of 3 is 2 and the other two elements 1 and 3 are odd 4 4 4 2 2 4 4 1 2 1 2 1 6 4 16 41 0 2 4 2 Coin Flips A standard object in combinatorics and probability is a sequence of coin flips A sequence of n flips is really just an element of Tn 2 where we treat 1 as Heads and 2 as Tails Binomial Theorem x y R n N n x y n X n k 0 k xk y n k We had you prove the Binomial Theorem in recitation last week and briefly mentioned it on Friday but I didn t really demonstrate its full power I actually neglected to mention that several of the results we proved in those Counting In Two Ways examples are proven directly by the Binomial Theorem For example we showed that 3n n n X n k X n n k 2 2 k k k 0 k 0 by counting ternary n tuples and partitioning based on how many positions are 0s Instead we can just apply the binomial theorem with x 2 and y 1 To get the other equality just use x 1 and y 2 instead Example Use x 1 and y 1 The Binomial Theorem tells us n n X X n k n k k n n n 1 1 1 1 1 0 0 k k k 0 k 0 Think about the terms of the sum whenever k is even the term is positive whenever k is odd the term 1 k is odd Thus we are positively counting all of the even sized subsets of n and negatively counting all of the odd sized subsets of n In total this sum is 0 This tells us there is an equal number of oddand even sized subsets of n We already proved this back on the Exam 2 Prep Questions by finding an explicit bijection between the two sets and finding its inverse This is just a different way of proving this same result Do you think it s better Worse I think it s neato to know there are multiple approaches to the same question Selections With Repetition Pirates Gold We have n indistinguishable identical gold coins to distribute amongst k distinguishable …