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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
A.B.C.Balbuena (UP-Math)   Introduction to Combinatorics   11.1.2008   10 / 10

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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