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### combinatorics

• 1. Introduction to Combinatorics A.Benedict Balbuena Institute of Mathematics, University of the Philippines in Diliman 11.1.2008 A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 1 / 10
• 2. Addition Rule Theorem If A1 , A2 , ..., An are disjoint sets (n < ∞, n ∈ N) then: |A1 ∪ A2 ∪ ... ∪ An | = |A1 | + |A2 | + ... + |An | Works only for disjoint sets One seat in a presidential working commitee is reserved for either a senator or a party-list representative. How many possible choices are there for the seat if there are 23 senators and 27 party-list representatives? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 2 / 10
• 3. Addition Rule Theorem If A1 , A2 , ..., An are disjoint sets (n < ∞, n ∈ N) then: |A1 ∪ A2 ∪ ... ∪ An | = |A1 | + |A2 | + ... + |An | Works only for disjoint sets One seat in a presidential working commitee is reserved for either a senator or a party-list representative. How many possible choices are there for the seat if there are 23 senators and 27 party-list representatives? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 2 / 10
• 4. Addition Rule Theorem If A1 , A2 , ..., An are disjoint sets (n < ∞, n ∈ N) then: |A1 ∪ A2 ∪ ... ∪ An | = |A1 | + |A2 | + ... + |An | Works only for disjoint sets One seat in a presidential working commitee is reserved for either a senator or a party-list representative. How many possible choices are there for the seat if there are 23 senators and 27 party-list representatives? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 2 / 10
• 5. Product Rule Recall: A × B = {(a, b)|a ∈ A, b ∈ B} Theorem If A1 , A2 , ..., An are sets (n < ∞, n ∈ N) then: |A1 × A2 × ... × An | = |A1 ||A2 |...|An | Interpret as number of ways to pick one item from A1 , one item from A2 , ..., one item from An A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 3 / 10
• 6. Product Rule Recall: A × B = {(a, b)|a ∈ A, b ∈ B} Theorem If A1 , A2 , ..., An are sets (n < ∞, n ∈ N) then: |A1 × A2 × ... × An | = |A1 ||A2 |...|An | Interpret as number of ways to pick one item from A1 , one item from A2 , ..., one item from An A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 3 / 10
• 7. Product Rule Recall: A × B = {(a, b)|a ∈ A, b ∈ B} Theorem If A1 , A2 , ..., An are sets (n < ∞, n ∈ N) then: |A1 × A2 × ... × An | = |A1 ||A2 |...|An | Interpret as number of ways to pick one item from A1 , one item from A2 , ..., one item from An A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 3 / 10
• 8. Examples 1 How many bit-strings are there of length n? 2 How many functions are there from a set of m elements to one with n elements? 3 How many possible mobile phone numbers are there in the Philippines? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 4 / 10
• 9. Inclusion-Exclusion Principle Theorem Let A, B be sets. Then: |A ∪ B| = |A| + |B| − |A ∩ B| Proof. By deﬁnitions of set difference and intersection, B = (A ∩ B) ∪ (B − A).Now, A ∩ B and B − A are disjoint sets.By the previous thm, |B| = |A ∩ B| + |B − A|.Since A and B − A are disjoint, |A ∪ B| = |A| + |B − A|.Replacing B − A, we have |A ∪ B| = |A| + |B| − |A ∩ B| How many bit strings of length of length 8 start with a 1 or end with 00? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 5 / 10
• 10. Inclusion-Exclusion Principle Theorem Let A, B be sets. Then: |A ∪ B| = |A| + |B| − |A ∩ B| Proof. By deﬁnitions of set difference and intersection, B = (A ∩ B) ∪ (B − A).Now, A ∩ B and B − A are disjoint sets.By the previous thm, |B| = |A ∩ B| + |B − A|.Since A and B − A are disjoint, |A ∪ B| = |A| + |B − A|.Replacing B − A, we have |A ∪ B| = |A| + |B| − |A ∩ B| How many bit strings of length of length 8 start with a 1 or end with 00? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 5 / 10
• 11. Inclusion-Exclusion Principle Theorem Let A, B be sets. Then: |A ∪ B| = |A| + |B| − |A ∩ B| Proof. By deﬁnitions of set difference and intersection, B = (A ∩ B) ∪ (B − A).Now, A ∩ B and B − A are disjoint sets.By the previous thm, |B| = |A ∩ B| + |B − A|.Since A and B − A are disjoint, |A ∪ B| = |A| + |B − A|.Replacing B − A, we have |A ∪ B| = |A| + |B| − |A ∩ B| How many bit strings of length of length 8 start with a 1 or end with 00? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 5 / 10
• 12. Inclusion-Exclusion Principle Theorem Let A, B be sets. Then: |A ∪ B| = |A| + |B| − |A ∩ B| Proof. By deﬁnitions of set difference and intersection, B = (A ∩ B) ∪ (B − A).Now, A ∩ B and B − A are disjoint sets.By the previous thm, |B| = |A ∩ B| + |B − A|.Since A and B − A are disjoint, |A ∪ B| = |A| + |B − A|.Replacing B − A, we have |A ∪ B| = |A| + |B| − |A ∩ B| How many bit strings of length of length 8 start with a 1 or end with 00? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 5 / 10
• 13. Inclusion-Exclusion Principle Theorem Let A, B be sets. Then: |A ∪ B| = |A| + |B| − |A ∩ B| Proof. By deﬁnitions of set difference and intersection, B = (A ∩ B) ∪ (B − A).Now, A ∩ B and B − A are disjoint sets.By the previous thm, |B| = |A ∩ B| + |B − A|.Since A and B − A are disjoint, |A ∪ B| = |A| + |B − A|.Replacing B − A, we have |A ∪ B| = |A| + |B| − |A ∩ B| How many bit strings of length of length 8 start with a 1 or end with 00? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 5 / 10
• 14. Inclusion-Exclusion Principle Theorem Let A, B be sets. Then: |A ∪ B| = |A| + |B| − |A ∩ B| Proof. By deﬁnitions of set difference and intersection, B = (A ∩ B) ∪ (B − A).Now, A ∩ B and B − A are disjoint sets.By the previous thm, |B| = |A ∩ B| + |B − A|.Since A and B − A are disjoint, |A ∪ B| = |A| + |B − A|.Replacing B − A, we have |A ∪ B| = |A| + |B| − |A ∩ B| How many bit strings of length of length 8 start with a 1 or end with 00? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 5 / 10
• 15. A password on a social networking site is six characters long, where each character is a letter or a digit. Each password must contain at least one digit. How many passwords are there? What if a password is six to eight chracters long? A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 6 / 10
• 16. Suppose we have 3 blue shirts, 2 red shirts, and 1 green shirt. We also have 2 gray pants and 3 brown pants. How many outﬁts are possible? (Two pieces of clothing with the same color are still considered distinct; assume that they have slightly different shades.) S := set of shirts P := set of pants. We form an outﬁt by picking one shirt and one pair of pants. By the Product Rule, there are (6)(5) = 30 outﬁts. A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 7 / 10
• 17. Suppose we have 3 blue shirts, 2 red shirts, and 1 green shirt. We also have 2 gray pants and 3 brown pants. How many outﬁts are possible? (Two pieces of clothing with the same color are still considered distinct; assume that they have slightly different shades.) S := set of shirts P := set of pants. We form an outﬁt by picking one shirt and one pair of pants. By the Product Rule, there are (6)(5) = 30 outﬁts. A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 7 / 10
• 18. What if gray pants go with only blue and red shirts and brown pants go with only green and red shirts. How many matching outﬁts are there? Count the number of matching outﬁts by counting the number of mismatching outﬁts. By the Product Rule, there are (2)(1) = 2 mismatching gray-green outﬁts and (3)(3) = 9 mismatching brown-blue outﬁts. Therefore, there are 3029 = 19 matching outﬁts. A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 8 / 10
• 19. How many ways can the ﬁve game NBA Finals Series be decided? (The series is decided when a team has three wins or three losses) A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 9 / 10
• 20. A.B.C.Balbuena (UP-Math) Introduction to Combinatorics 11.1.2008 10 / 10
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