Section 3: Counting. Elements in Disjoint Sets
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1 Section 3: Counting Elements in Disjoint Sets In the last section, we looked at counting events that branch from one state to the next. These events we modelled using possibility trees and obtained counts using the Multiplication Rule. In this section, we will learn how to count elements in the union, difference, or intersection of two sets using the Addition Rule. We will also look at counting when set operations are acting upon subset structures.
2 6.3.2 The Addition Rule Theorem: Suppose a finite set A is partitioned by the collection {A 1, A 2,..., A n }. Then: n(a) = n(a 1 ) + n(a 2 ) n(a n ). How many binary strings of length up to and including 4 are there? Answer: Γ 4 = Σ 0 Σ 1 Σ 2 Σ 3 Σ 4, and Σ i Σ j = if i j, so Γ 4 = Σ 0 + Σ 1 + Σ 2 + Σ 3 + Σ 4 = = = 31.
3 6.3.3 Another Example Let D be set of 3-digit integers (from 100 to 999) which are divisible by 5. How big is D? Solution: We know an integer is divisible by 5 if it ends in a 0 or 5. Let s call a typical 3-digit integer abc, where a is one of {1,2,3,4,5,6,7,8,9}, and b is one of {0,1,2,3,4,5,6,7,8,9}. Now, we partition D into those integers of the form ab0 and those of the form ab5. The first set has 9 10 elements and the second set has 9 10 elements, hence D has 180 elements.
4 6.3.4 License Plates We have already used the addition rule without realizing it in the last section. When we are counting the number of license plates with certain properties and allow a particular character to be either a digit or a letter, we have applied the addition rule: The slot drawing (Digit or Letter) (Digit or Letter)... gives us the counts ( ) ( )...
5 6.3.5 The Difference Rule Consider the following: A 1 A 2 so A = A 1 + A 2. Now, in the addition rule, we know A 1 and A 2 and use this to find A. But if we know A and either A 1 or A 2, then we get: Difference Rule: If A 1 and A 2 partition A, then A 2 = A A 1 and A 1 = A A 2. A
6 Counting Things Without a Condition How many 3-digit integers are not divisible by 5? Before, we saw that there are digit integers which are divisible by 5, hence: {all 3-digit integers not divisible by 5} = {all 3-digit integers} {all 3-digit integers divisible by 5}. Therefore there are ( ) 180 = digit integers which are not divisible by
7 Another Complement Example How many ways can 4 couples sit around a circular table if one couple cannot be seated next to each other? Solution: We see that {seatings w/ couple apart} = {all seatings} {seating w/ couple together} Label the people abcdefgh and suppose couple ab cannot sit together. Thus we want to count how to seat Xcdefgh around a table, then repeat it since X represents both the ab and ba orderings. Answer = 7! 6! 6! = 7! 2 6!
8 A Probability Example Suppose there are 10,000 total ways to carry out an experiment with 2,123 of those experiments leading to success. What is the probability the experiment will fail? Solution: Starting from the sample space, we get: {all experiments} = {successes} + {failures}, Dividing boths sides by the LHS, we see that: 1 = P(success) + P(failure), so P(failure) = 1 P(success) = = We find there is a 78.77% probability of failure.
9 Inclusion/Exclusion Rule So far, our counting under the addition and subtraction rules have required that the files in question are disjoint This is an unreasonable requirement in general. How do we account for the aggregrate knowing the parts that make it up? A A B B A B Inclusion/Exclusion Rule: For any sets A and B, A B = A + B A B.
10 Inclusion/Exclusion Example How many 6-character license plates begin with A or end with 9? Effectively, we want to count all plates of the forms Axxxxx or xxxxx9, but recognizing that these individual cases overlap for plates of the form Axxxx9. Now {Axxxxx} = 36 5, and {xxxxx9} = 36 5, and {Axxxx9} = 36 4, hence the total number of plates is
11 What About For 3 Sets? Using the labeling in the above drawing, we see that A = , B = , 3 and C = Thus A B C = A + B + C ( ) A B A C B C ( ) + A B C (+7) =
12 Inclusion/Exclusion Rule for 3 Sets If A, B, and C are sets, then: A B C = A + B + C A B A C B C + A B C. Example: The CS department teaches Algol, Basic, and C languages. Suppose: 22 students study Algol; 30 study Basic; 42 study C; 12 study Algol and Basic; 18 study Basic and C; 16 study Algol and C; 9 study all three. How many students study a language? Answer: 57 =
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