1 / 12

Section 6.5

Section 6.5. Inclusion/Exclusion. Finding the number of elements in the union of 2 sets. From set theory, we know that the number of elements in the union of 2 sets is the sum of the number of elements in each set minus the number of elements in the intersection of the 2 sets:

daktari
Download Presentation

Section 6.5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Section 6.5 Inclusion/Exclusion

  2. Finding the number of elements in the union of 2 sets • From set theory, we know that the number of elements in the union of 2 sets is the sum of the number of elements in each set minus the number of elements in the intersection of the 2 sets: |A  B| = |A| + |B| - |A  B|

  3. Example 1 • A discrete math class consists of 4 students taking Software Design, 3 students taking CS2, 2 students taking neither, and 1 student taking both. How many students are in the class? • Let |A| = # in SD, |B| = # in CS2, |C| = # in neither • So |AB| = # taking both and |ABC| = # in discrete • |A  B  C| = |A| + |B| + |C| - |A  B| = 4+3+2-1=8

  4. Example 2 • How many positive integers not exceeding 100 are divisible by 2 or 5? • |A| = # divisible by 2 = 100/2 = 50 • |B| = # divisible by 5 = 100/5 = 20 • |A  B| = # divisible by both; since they are mutually prime, this is the numbers divisible by 2*5 = 100/10 = 10 • So |A  B| = 50 + 20 - 10 = 60

  5. Example 3 • We can use similar means to find the number of elements outside the union of 2 sets: • A recent survey found that 96% of U.S. households have at least one television, 98% have phone service, and 95% have both; what percentage of households have neither?

  6. Solution for example 3 • Let |A| = % of households with TV (96) and |B| = % of households with phone service (98) • We know that 95% have both; this is |AB| • The total number of households that have either TV or phone service, |AB| is: |A| + |B| - |AB| = 96+98-95 = 99 • The total number of households is 100%, so the number that have neither TV nor phone is 100-99, or 1%

  7. Finding the number of elements in the union of 3 sets • The sum |A| + |B| + |C| counts the number of elements in one set once, the number in 2 sets twice, and the number in all 3 sets 3 times • Subtracting the number of elements in any pair of the sets eliminates the double counting: |A|+|B|+|C| - |AB| - |AC| - |BC| • But this also eliminates all elements appearing in all 3; so we add those elements back in: |A|+|B|+|C| - |AB| - |AC| - |BC| + |ABC|

  8. Example 4 • Suppose there are 2504 computer science majors at a school; of these: • 1876 have taken C++, 999 have taken Java, 345 have taken C and • 876 have taken both Java and C++, 231 have taken C++ and C, and 290 have taken Java and C • How many CS majors have not taken any of these languages?

  9. Example 4 • Let |J| be the number who have taken Java (999), |P| be the number who have taken C++ (1876) and |C| the number who have taken C (345) • Then: |JPC| = |J|+|P|+|C| - |JP| - |JC| - |PC| + |JPC| 999+1876+345 - 876 - 231 - 290 + 189 = 2012 • So 2504 - 2012 or 492 have taken none of these languages

  10. Principle of Inclusion/Exclusion Let A1, A2, … , An be finite sets; then |A1 A2  …  An| =  |Ai| -  |Ai Aj| + 1in1i<jn  |Ai Aj Ak| - … + 1i<j<kn (-1)n+1|A1 A2 …  An|

  11. Proof We want to show that each element in the union of the sets is counted exactly once; suppose a is a member of exactly r of the sets A1 … An where 1  r  n. This element is counted: C(r,1) times by |Ai| and C(r,2) times by |Ai  Aj| and, in general, C(r,m) times by summation involving m of the sets So element a is counted C(r,1)-C(r,2)+C(r,3)- … +C(r,r) times

  12. Proof continued Recall the binomial theorem: (x+y)n =  C(n,j)xn-jyj (as j goes from 0 to n) We can show from this theorem that (-1)kC(n,k) = 0 (as k goes from 0 to n); thus C(r,1) - C(r,2) + C(r,3)- … +(-1)rC(r,r) = 0 Changing the exponent of -1 in the last term to r+1 gives us: C(r,1) - C(r,2) + C(r,3)- … +(-1)r+1C(r,r) = 1 So each element in the union is counted exactly once.

More Related