Counting Elements of Disjoint Sets: The Addition Rule Lecture 26 Section 6.3 Fri, Feb 25, 2005
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|.
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|.
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).
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|.
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?
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|.
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 = 260530, so there are 61 31 = 1891 divisors of 2030.
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, 1681 + 1891 – 1271 = 2301 numbers divide either 1040 or 2030.
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.”
Proof, continued Proof: |A B C| = |A B| + |C| – |(A B) C|.
Proof, continued Proof: |A B C| = |A B| + |C| – |(A B) C| = |A| + |B| – |A B| + |C| – |(A B) C|.
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)|.
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)|.
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|.
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|
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|
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).
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|.