Lecture 34 21 127 Concepts of Math 11 26 2012 Administrivia Exam 3 Prep Questions are posted Exam 3 is on Friday HW 5 grades are now posted under Homework 5 B on CMG Exam 2 grades are uploaded Please check your grades and make sure they are correct Email me about any errors and or bring your homework exam to class and hand it to me so I can investigate Office hours this week the usual on MTW plus Thursday from 12 2pm and 4 6pm There are only 5 lectures left including this one and not including Friday s exam and only 4 recitations You will have a short homework due next Thursday I recognize it s the last week and it s right after an exam but you need to work on some problems related to this new material Email me ASAP about scheduling conflicts extra time for the final exam which is on Thursday December 13 from 8 30 11 30am I will notify you of office hours and review sessions for the final later on I will also post Prep Questions for the final as soon as possible sometime in the middle of next week so you have over a week to work on them Summarizing Counting Formulas Say we have n objects and we want to select k of them How many ways can we do this The answer depends on two questions Are repeats allowed Does order distinguish the outcome Each of these questions can be answer with Yes or No and each of the four ways to answer them yields a different formulation of the original question Yes No Yes No nk n n k n k k n 1 k Vertical answers Does order distinguish Horizontal answers Are repeats allowed Note Sometimes the roles of n and k are reversed in a problem Be careful about this In particular we usually say there are n gold pieces and k pirates in a counting problem however this is equivalent to saying there are k types of objects and we are selecting n of them which is the opposite of the situation represented in the table 1 Selections With Repetition Pirates Gold Let s say we have n 20 pieces of gold to distribute amongst k 3 pirates Captains Redbeard Khair ad Din an Ottoman Blackbeard Edward Teach an Englishman and Kidd a Scotsman Let s figure out how many ways there are to distribute the gold under certain conditions How many ways are there total This is like selecting 20 objects from 3 types with repetition allowed Whenever we select a pirate that means we give him a piece of gold By the above selection formula there are 20 3 1 22 22 21 231 20 20 2 How many ways ensure every pirate gets at least 2 pieces Let s just give everyone two pieces of gold right away Then we have 20 6 14 pieces of gold left to distribute amongst all 3 pirates so there are 14 2 16 16 15 120 2 14 14 How many ways ensure Redbeard and Blackbeard get at least 2 and Kidd gets at least 6 Just like the last one let s give Red Blackbeard 2 and Kidd 6 pieces This leaves us with 20 4 6 10 pieces left to distribute amongst all 3 pirates so there are 10 2 12 12 11 66 10 10 2 How many ways ensure Redbeard and Blackbeard get at least 2 while Kidd gets no more than 2 There are a couple of ways to approach this one i Let s establish cases based on whether Kidd gets 0 or 1 or 2 golds In each case we will give Red Blackbeard 2 pieces right away and then give Kidd the corresponding amount 0 or 1 or 2 This leaves us with 16 or 15 or 14 pieces left to distribute only amongst the first two pirates so there are 16 1 15 1 14 1 16 15 14 45 16 15 14 ii Let s consider all of the ways to ensure Red Blackbeard get at least 2 golds and then remove from this the number of ways where Kidd gets too many i e at least 3 golds If we give Red Blackbeard 2 golds then we have 16 pieces left to distribute amongst all 3 of the each 18 17 153 ways pirates so there are 16 2 18 16 16 2 Then if we give Red Blackbeard 2 golds each Kidd 3 and then distribute the remaining 13 amongst and 15 15 14 all 3 of the pirates there are 13 2 105 Thus in total there are 13 13 2 18 15 153 105 48 16 13 2 How many ways ensure no pirate gets more than 9 pieces How would you approach this one We ll leave it open ended I really only asked this to see what you could innovate My approach would be to idenify the possible distributions we could have that is I would identify the number of ways to have a sum of 3 numbers make 20 without having any of the numbers larger than 9 which are 992 983 974 965 884 875 866 774 There are other reasonable approaches as well See what you can come up with By the way how did we prove the formula n k 1 represents the total number of ways Do you remember k 1 the gold and dividers argument Keep that counting in two ways argument in your head rather than memorizing the formula You can always reconstruct it Indistinguishable Dice Consider rolling 20 indistinguishable dice How many total outcomes are there How many have each number appearing at least twice How many have at most three 6s How many have at least four 6s How many have no more than 5 appearances of any number We ll leave these questions to you as exercises They use similar counting arguments to the above examples though Think of finding the number of solutions to x1 x2 x3 x4 x5 x6 20 where xi is the number of dice whose face shows i Inclusion Exclusion The general idea here is that we want to count the elements of a universal set that are not elements of certain subsets of that universe So we can count the whole universe then remove the elements of each subset but then add back in the elements in their intersections but then remove again the double intersections Essentially we keep fixing an over undercount until it is correct By the way throughout this section we are of course referring to only finite sets Consider A1 U Then U A1 U A1 Note This only works because A1 U The claim is false otherwise Consider A1 U and A2 U Then U A1 A 2 U A1 A2 A1 A2 Think about why this works Consider an element that we want to count on the left hand side It is an element x U such that x A1 and x A2 Then how many times is x counted on the right hand side …