Section 6 5 InclusionExclusion 1 Finding the number

Section 6. 5 Inclusion/Exclusion 1

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| 2

Example 1 • A discrete math class consists of 4 students taking Software Design, 3 students taking CS 2, 2 students taking neither, and 1 student taking both. How many students are in the class? – Let |A| = # in SD, |B| = # in CS 2, |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 3

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 4

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? 5

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% 6

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| 7

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? 8

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 9

Principle of Inclusion/Exclusion Let A 1, A 2, … , An be finite sets; then |A 1 A 2 … An| = |Ai| - |Ai Aj| + 1 i n 1 i<j n |Ai Aj Ak| - … + 1 i<j<k n (-1)n+1|A 1 A 2 … An| 10

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 A 1 … 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 11

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)k. C(n, k) = 0 (as k goes from 0 to n); thus C(r, 1) - C(r, 2) + C(r, 3)- … +(-1)r. C(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+1 C(r, r) = 1 So each element in the union is counted exactly once. 12