Inclusion/Exclusion

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Section 6.5

• Inclusion/Exclusion

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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

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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
  432-18

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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
  25 ?100/10? 10
• So A ? B 50 20 - 10 60

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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?

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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 9698-95 99
• The total number of households is 100, so the
  number that have neither TV nor phone is 100-99,
  or 1

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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
• ABC - A?B - A?C - B?C
• But this also eliminates all elements appearing
  in all 3 so we add those elements back in
• ABC - A?B - A?C - B?C A?B?C

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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?

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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 JPC - J?P - J?C - P?C
  J?P?C
• 9991876345 - 876 - 231 - 290 189 2012
• So 2504 - 2012 or 492 have taken none of these
  languages

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Principle of Inclusion/Exclusion

• Let A1, A2, , An be finite sets then
• A1 ? A2 ? ? An
• ? Ai - ? Ai ? Aj
• 1?i?n 1?iltj?n
• ? Ai ? Aj ? Ak -
• 1?iltjltk?n
• (-1)n1A1 ? A2 ? ? An

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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
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Proof continued
Recall the binomial theorem (xy)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 r1 gives us C(r,1) -
C(r,2) C(r,3)- (-1)r1C(r,r) 1 So each
element in the union is counted exactly once.