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Counting Elements of Disjoint Sets: The Addition Rule

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1 Counting Elements of Disjoint Sets: The Addition Rule
Lecture 26 Section 6.3 Fri, Feb 25, 2005

2 Counting Elements in Disjoint Sets
Theorem: Let {A1, …, An} be a partition of a set A. Then |A| = |A1| + … + |An|. Corollary: Let {A1, …, An} be a collection of pairwise disjoint finite sets. Then |A1  …  An| = |A1| + … + |An|.

3 Counting Elements in Subsets
Theorem: Let A and B be finite sets with B  A. Then |A – B| = |A| – |B|. Proof: {B, A – B} is a partition of A. Therefore, |B| + |A – B| = |A|. So, |A – B| = |A| – |B|.

4 Counting Elements in Subsets
Corollary: Let A and B be finite sets with B  A. Then |B| = |A| – |A – B|. Corollary: Let S be the sample space of an experiment and let E be an event. Then P(Ec) = 1 – P(E).

5 Counting Elements in Unions of Sets
Theorem: Let A and B be any finite sets. Then |A  B| = |A| + |B| – |A  B|. Proof: (A  B) – B = A – (A  B). Furthermore, B  A  B and A  B  A. Therefore, |A  B| – |B| = |A| – |A  B|. So, |A  B| = |A| + |B| – |A  B|.

6 P(A  B) = P(A) + P(B) – P(A  B).
Probability Corollary: Given an experiment, let A and B be two events. Then P(A  B) = P(A) + P(B) – P(A  B). Example: Draw two cards from a deck. What is the probability that at least one of them is black?

7 Putnam Problem A-1 (1983) How many positive integers n are there such that n is an exact divisor of at least one of the numbers 1040, 2030? Let A = {n | n divides 1040}. Let B = {n | n divides 2030}. Then |A  B| = |A| + |B| – |A  B|.

8 Putnam Problem A-1 (1983) Prime factorization: 1040 = 240  540.
Therefore, n | 1040 if and only if n = 2a5b where 0  a  40 and 0  b  40. There are 41  41 = 1681 such numbers. Similarly, 2030 = , so there are 61  31 = 1891 divisors of 2030.

9 Putnam Problem A-1 (1983) Finally, an integer is in A  B if it divides both 1040 and 2030. That means that it divides the gcd of 1040 and 2030. The gcd of 240  540 and 260  530 is 240  530. Therefore, there are 41  31 = 1271 such numbers. Thus, – 1271 = 2301 numbers divide either 1040 or 2030.

10 Number of Elements in the Union of Three Sets
Theorem: Let A, B, and C be any three finite sets. Then |A  B  C| = |A| + |B| + |C| – |A  B| – |A  C| – |B  C| + |A  B  C|. The pattern is to add “one at a time”, subtract “two at a time”, and add “three at a time.”

11 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|.

12 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|
= |A| + |B| – |A  B| + |C| – |(A  B)  C|.

13 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|
= |A| + |B| – |A  B| + |C| – |(A  B)  C| = |A| + |B| – |A  B| + |C| – |(A  C)  (B  C)|.

14 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|
= |A| + |B| – |A  B| + |C| – |(A  B)  C| = |A| + |B| – |A  B| + |C| – |(A  C)  (B  C)| = |A| + |B| – |A  B| + |C| – |A  C| – |B  C| + |(A  C)  (B  C)|.

15 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|
= |A| + |B| – |A  B| + |C| – |(A  B)  C| = |A| + |B| – |A  B| + |C| – |(A  C)  (B  C)| = |A| + |B| – |A  B| + |C| – |A  C| – |B  C| + |(A  C)  (B  C)| + |A  B  C|.

16 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|
= |A| + |B| – |A  B| + |C| – |(A  B)  C| = |A| + |B| – |A  B| + |C| – |(A  C)  (B  C)| = |A| + |B| – |A  B| + |C| – |A  C| – |B  C| + |(A  C)  (B  C)| + |A  B  C|. = |A| + |B| + |C| – |A  B| – |A  C| – |B  C|

17 Proof, continued Proof: |A  B  C| = |A  B| + |C| – |(A  B)  C|
= |A| + |B| – |A  B| + |C| – |(A  B)  C| = |A| + |B| – |A  B| + |C| – |(A  C)  (B  C)| = |A| + |B| – |A  B| + |C| – |A  C| – |B  C| + |(A  C)  (B  C)| + |A  B  C|. = |A| + |B| + |C| – |A  B| – |A  C| – |B  C|

18 Number of Elements in the Union of Three Sets
Corollary: Let A, B, and C be any three events. Then P(A  B  C) = P(A) + P(B) + P(C) – P(A  B) – P(A  C) – P(B  C) + P(A  B  C).

19 The Inclusion/Exclusion Rule
Theorem: Let A1, …, An be finite sets. Then |A1  …  An| = i |Ai| – i j > i |Ai  Aj| + i j > i k > j |Ai  Aj  Ak| :  |A1  …  An|.


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