Download Section 5.5

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| = 4+3+2-1=8
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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
2*5 = 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| = 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%
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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:
|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|
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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| = |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
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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i<jn
|Ai  Aj  Ak| - … +
1i<j<kn
(-1)n+1|A1  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:
(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.
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