Presentation on theme: "Counting Techniques: Permutations of Selected Elements Addition Rule, Difference Rule, Inclusion/Exclusion Rule."— Presentation transcript:
1Counting Techniques: Permutations of Selected Elements Addition Rule, Difference Rule, Inclusion/Exclusion Rule
2Permutations of Selected Elements Typical situation: A chairman, a secretary and a treasurer are to be chosen in a committee of 7 people.Question: In how many different ways can it be done?Definition: An r-permutation of a set S of n elements is an ordered selection of r elements taken from S.The number of all r-permutations of a set of n elements is denoted P(n,r) .In the example above we want to find P(7,3).
3How to compute P(n,r) Theorem: P(n,r) = n(n-1)(n-2)…(n-r+1) or, equivalently,Proof:Forming an r-permutation of a set of n-elementsis an r-step operation: Step 1: Choose the 1st element ( n different ways). Step 2: Choose the 2nd element ( n-1 different ways).… Step r: Choose the rth element (n-r+1 different ways).Based on the multiplication rule,the number of r-permutations is n∙(n-1)∙…∙(n-r+1) .
4Examples of r-permutations 1. Choosing a chairman, a secretary and a treasurer among 7 people:P(7,3) = 7∙6∙5 = 210 .2. Suppose Jim is already chosen to be the secretary.Q: How many ways a chairman and a treasurer can be chosen?A: P(6,2) = 6∙5 = 30 .3. In an instance of the Traveling Salesman Problem,the total number of cities = 10;this time the salesman is supposed to visit only 4 cities (including the home city).Q: How many different tours are possible?A: P(9,3) = 9∙8∙7 = 504 .
5The Addition Rule Suppose a finite set A equals the union of k distinct mutually disjoint subsets A1, A2, …, Ak .Then n(A) = n(A1) + n(A2) + … + n(Ak)Example: How many integers from 1 through 999do not have any repeated digits?Solution: Let A = integers from 1 to 999not having repeated digits.Partition A into 3 sets:A1=one-digit integers not having repeated digits;A2=two-digit integers not having repeated digits;A3=three-digit integers not having repeated digits.Then n(A)=n(A1)+n(A2)+n(A3) (by the addition rule)= 9 + 9∙9 +9∙9∙8 = 738. (by the multipl. rule)
6The Difference Rule If A is a finite set and B is a subset of A, then n(A-B) = n(A) – n(B) .Example: Assume that any seven digits can be usedto form a telephone number.Q: How many seven-digit phone numbershave at least one repeated digit?Let A = the set of all possible 7-digit phone numbers;B = the set of 7-digit numbers without repetition.Note that BA .Then A-B is the set of 7-digit numbers with repetition.n(A-B) = n(A) – n(B) (by the difference rule)= 107 – P(10,7) (by the multiplication rule)= 107 – 10! / 3! = 10,000,000 – 3,628,800/6 == 10,000,000 – 604,800 = 9,395,200
7The Inclusion/Exclusion Rule for Two or Three Sets If A, B and C are finite sets then n(A B) = n(A) + n(B) – n(A B) n(A B C) = n(A) + n(B) + n(C)- n(A B) – n(A C) – n(B C)+ n(A B C)ABABC
8Example on Inclusion/Exclusion Rule (2 sets) Question: How many integers from 1 through 100are multiples of 4 or multiples of 6 ?Solution: Let A=the set of integers from 1 through which are multiples of 4;B = the set of integers from 1 through 100which are multiples of 6.Then we want to find n(A B).First note that A B is the set of integersfrom 1 through 100 which are multiples of 12 .n(A B) = n(A) + n(B) - n(A B) (by incl./excl. rule)= – 8 = 33 (by counting the elementsof the three lists)
9Example on Inclusion/Exclusion Rule (3 sets) 3 headache drugs – A,B, and C – were tested on 40 subjects. The results of tests:23 reported relief from drug A;18 reported relief from drug B;31 reported relief from drug C;11 reported relief from both drugs A and B;19 reported relief from both drugs A and C;14 reported relief from both drugs B and C;37 reported relief from at least one of the drugs.Questions:1) How many people got relief from none of the drugs?2) How many people got relief from all 3 drugs?3) How many people got relief from A only?
10Example on Inclusion/Exclusion Rule (3 sets) We are given: n(A)=23, n(B)=18, n(C)=31,n(A B)=11, n(A C)=19, n(B C)=14 ,n(S)=40, n(A B C)=37Q1) How many people got relief from none of the drugs?By difference rule,n((A B C)c ) = n(S) – n(A B C) = = 3SABC
11Example on Inclusion/Exclusion Rule (3 sets) Q2) How many people got relief from all 3 drugs?By inclusion/exclusion rule:n(A B C) = n(A B C)- n(A) - n(B) - n(C)+ n(A B) + n(A C) + n(B C)= 37 – 23 – 18 – = 9Q3) How many people got relief from A only?n(A – (B C)) (by inclusion/exclusion rule)= n(A) – n(A B) - n(A C) + n(A B C)= 23 – 11 – = 2