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Section 6.5Section 6.5

Inclusion/Exclusion

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Finding the number of elements Finding the number of elements in the union of 2 setsin 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 1Example 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 |AB| = # taking both and |ABC| = # in

discrete – |A B C| = |A| + |B| + |C| - |A B| = 4+3+2-1=8

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Example 2Example 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 3Example 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 3Solution 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 |AB|• The total number of households that have either

TV or phone service, |AB| is:|A| + |B| - |AB| = 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 Finding the number of elements in the union of 3 setsin 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| - |AB| - |AC| - |BC|

• But this also eliminates all elements appearing in all 3; so we add those elements back in:

|A|+|B|+|C| - |AB| - |AC| - |BC| + |ABC|

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Example 4Example 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 4Example 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:|JPC| = |J|+|P|+|C| - |JP| - |JC| - |PC| + |JPC|

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/ExclusionPrinciple of Inclusion/Exclusion

Let A1, A2, … , An be finite sets; then

|A1 A2 … An| =

|Ai| - |Ai Aj| + 1in 1i<jn |Ai Aj Ak| - … + 1i<j<kn

(-1)n+1|A1 A2 … An|

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ProofProofWe 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| andC(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 continuedProof 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); thusC(r,1) - C(r,2) + C(r,3)- … +(-1)rC(r,r) = 0

Changing the exponent of -1 in the last term to r+1gives us:

C(r,1) - C(r,2) + C(r,3)- … +(-1)r+1C(r,r) = 1So each element in the union is counted exactly once.