Math 55 - Spring 2004 - Lecture notes # 14 - March 9 (Tuesday)Read Chapter 4Goals for Today: Continue counting principlesTree diagramsPigeonhole principlePermutationsCombinationsHomework:1) (Based on a recent 1st grade homework assignment froma local school.) Perhaps this is covered later?1.1) Suppose you can have 9 pieces of fruit, which canbe either apples or oranges. How many ways can youhave n pieces of fruit (for example, you could have4 apples and 5 oranges, or 0 apples and 9 oranges, etc.)1.2) Answer 1.1) for n pieces of fruit.1.3) Suppose you can have 9 pieces of fruit, which can beapples, oranges or pears. How many ways can you have9 pieces of fruit (actual 1st grade assignment).1.4) Answer 1.3) for n pieces of fruit.1.5) Suppose you can have n pieces of fruit, and there arem different kinds of fruit. How many ways can you haven pieces of fruit?2) (Also based on a recent 1st grade homework assignment.)Suppose you can make shapes from putting together identicalequilateral triangles edge-to-edge. Triangles must not overlap,and if they are adjacent, they must line up along an entire edge.In the picture below, the leftmost shape is ok, and the othertwo are not. Shapes must be connected (i.e. you can get fromany triangle to any other triangle by moving across adjacentedges).How many different shapes using 2, 3, 4, 5, 6 and 7 trianglesare there? Shapes are considered the same if one shape canbe slid or rotated (but not turned over) to lie exactly on topof the other shape. (The 6 triangle problem was the actual1st grade assignment. See if you can do 7.)13) 4.1- 20, 424) 4.2- 10, 145) 4.3- 18, 246) 4.4- 10, 387) 4.5- 20, 54Review Counting Principles1) The Sum Rule:If S1 and S2 are disjoint sets, then the number of members ofS1 U S2 is |S1 U S2| = |S1| + |S2|2) Inclusion-Exclusion Principle:If S1 and S2 are arbitrary sets, then|S1 U S2| = |S1| + |S2| - |S1 inter S2|3) The Product Rule:If S1 and S2 are sets, and S1 x S2 ={(s1,s2) : s1 in S1 and s2 in S2}is the Cartesian product of S1 and S2, then |S1 x S2| = |S1| x |S2|If S1, S2, ... , Sk are sets, S1 x S2 x ... Sk the Cartesian product,then |S1 x ... x Sk| = |S1| x ... x |Sk|4) Tree diagramsEX: How many bit strings of length 4 without two consecutive zeros?Ans: 8; Enumerate all possibilities in a tree:Legal bit strings0 - 1 0101/ / 0 01100 - 1 - 1 - 1 0111/ / 0 1010/ 0 - 1 - 1 1011/ / 0 - 1 1101/ / / / 0 1110x -----1 - 1 - 1 - 1 1111Bit#1234Tree Diagrams (formally) Draw a tree where the children of each noderepresent all the possible values of the next entryEX: How many 2-out-of-3 game playoffs are there? Ans: 62Winner sequence Winnerbbb b/ / b bab bb-a-a baa a/ / b abb b/ / b - a aba ax--a-a aa a(5) Pigeonhole Principle:If k+1 or more objects (pigeons) are placed in k boxes (holes),then at least one box contains 2 or more objects.(proof by contradition: if each box had at most one object,there would only be k or fewer objects, a contradiction)EX: In any group of 27 English words, at 2 begin with the sameletter, since there are only 26 letters.ASK&WAIT: How large a group of people do you need to be sure thattwo of them have the same first and last initials?ASK&WAIT: How many times do you have to shuffle a deck of cards, tobe sure that the cards are in exactly the same order at least twice?(6) Generalized Pigeonhole Principle:If N or more objects (pigeons) are placed in k boxes (holes),then at least one box contains ceiling(N/k) or more objects.(proof by contradition: if each box had at most ceiling(N/K)-1 objects,there would be at mostk*(ceiling(N/k) -1) < k*((N/k +1) - 1) = N objects, a contradiction)EX: N=k+1 implies ceiling(N/k)=ceiling((k+1)/k)=2, usual Pigeonhole principleASK&WAIT: There are 200 students enrolled in Math 55. How many have toreceive the same letter grade (A,B,C,D,F)?EX: Given any set S of n+1 positive integers less than or equal to 2*n,then one of them must divide another one:(ex: n=5, if S={2 3 5 7 9}, then 3|9)Proof: Let S = {a(1),a(2),...,a(n+1)}. Write a(i) = 2^(k(i)) * q(i),where q(i) is odd. So {q(1),...,q(n+1)} is a set of positiveodd integers from 1 to 2n-1, of which there are only n, namely1,3,5,...,2n-1. So by the pigeonhole principle, q(i)=q(j)=qfor some i and j. Thus a(i) = 2^(k(i)) * q anda(j) = 2^(k(j)) * q. If k(i)>k(j) then a(j)|a(i), else a(i)|a(j).3ASK&WAIT: Assuming California has 36M people, how many of themhave the same 3 initials and were born on the same day of the samemonth (but not necessarily in the same year)?For example, Arnold B. Casey (ABC), born 29 Feb 1955 andAbigail B. Chen (ABC), born 29 Feb 1990EX: Suppose you have n>1 computers, each of which chooses m othercomputers to which to establish a network connection. Computerscan communicate in either direction over such a connection,and can forward messages from one computer to another.What is the smallest value of m such that there is a guaranteedpath from any computer to any other computer, no matter where theconnections go?Exs: n=2 (m=1), n=3 (m=1), n=4 (m=2)Solution: note that the answer <= n-1, since with n-1, every computeris directly connected to every other computer.We answer by asking what is the largest k such that each computercan be connected to k others, such that there is no path betweensome pair of computers, then m=k+1Let N={all computers}. Suppose computer 1 can’t reach 2.Let N1 be all the computers 1 can reach, and N2 be all ones itcannot reach (such as 2). Then N={all computers} = N1 U N2.The most number of edges you can have just connecting computersinside Ni is |Ni|-1, which gives a direct connection from everycomputer inside Ni to every other. Any more edges would connectsome computer in Ni to some outside. Thus the max number ofedges which can keep N1 and N2 unconnected ismin(|N1|-1,|N2|-1) = min(|N1|,|N2|)-1. We want to pick |N1| to makethis as large as possible, because this will tell us the largestnumber k of connections per computer we can have and not becompletely connected. The largest min(|N1|,|N2|)-1 can be isk= min(floor(n/2),ceiling(n/2))-1 = floor(n/2)-1 and ifwe have m=k+1=floor(n/2) edges, we are guaranateed a connection.EX: Suppose you have a group of 6 people, where any 2 people areeither friends or enemies. Show that either you have 3 mutualfriends, or 3 mutual enemies, in the group.Proof: Take person A. Of the remaing 5 people, either at least 3are friends of A, or at least 3 are enemies.Suppose first that at least 3 are friends of A. If anypair of these (say B and C) are friends, then