1 / 24

Ch. 5: Counting

Ch. 5: Counting. 5.1: The Basics of Counting. Intro Example #1: If we have a class of 6 math majors and 7 CS majors (with no double majors) A) In how many ways could we pick one representative from the class? B) In how many ways could we pick 1 math and 1 CS major?

hanley
Download Presentation

Ch. 5: Counting

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Ch. 5: Counting

  2. 5.1: The Basics of Counting • Intro Example #1: If we have a class of 6 math majors and 7 CS majors (with no double majors) • A) In how many ways could we pick one representative from the class? • B) In how many ways could we pick 1 math and 1 CS major? • Intro Example #2: If a class has 10 math and 13 CS majors, and 4 of these are joint majors, how many are in the class?

  3. Summary of Basic Counting Techniques • Ex 1A uses the SUM rule • 6+7=13 • Ex 1B uses the PRODUCT rule • 6*7=42 • Ex 2 uses INCLUSION/EXCLUSION • 10+13-4=19

  4. Multiplication Problems 1.      At a restaurant, you have a choice of main dish (beef, chicken, fish, vegetarian), vegetable (broccoli, corn), potato (baked, fries), and dessert (chocolate, strawberry).  LIST all possible choices. 2.      A teacher wishes to make all possible different answer keys to a multiple choice quiz.  How many possible different answer keys could there be if there are 3 questions that each have 4 choices (A,B,C,D).  LIST them all. 3.      What if there were 20 multiple choice questions with 5 choices each? Explain (don’t list). 4.      With 9 baseball players on a team, how many different batting orders exist?

  5. Answers 1.      At a restaurant, you have a choice of main dish (beef, chicken, fish, vegetarian), vegetable (broccoli, corn), potato (baked, fries), and dessert (chocolate, strawberry).  LIST all possible choices. main vegetable potato dessert Beef broc baked chocolate Beef broc baked strawb Beef broc fries chocolate … 4*2*2*2=32

  6. Answers 2.      A teacher wishes to make all possible different answer keys to a multiple choice quiz.  How many possible different answer keys could there be if there are 3 questions that each have 4 choices (A,B,C,D).  LIST them all. 4*4*4=64 3.      What if there were 20 multiple choice questions with 5 choices each? Explain (don’t list). 5^20 4.      With 9 baseball players on a team, how many different batting orders exist? 9! = 362,880

  7. Permutation Examples 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and we wish to elect a president and vice-president, LIST all of the different ways that this is possible. 2. From these 4 people (Anne, Bob, Cindy, Dave), we wish to elect a president, vice-president, and treasurer. LIST all of the different ways that this is possible. 

  8. Answers 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and we wish to elect a president and vice-president, LIST all of the different ways that this is possible. AB BA CA DA AC BC CB DB AD BD CD DC 4*3=12 or 4P2 = 12

  9. Answers 2. From these 4 people (Anne, Bob, Cindy, Dave), we wish to elect a president, vice-president, and treasurer. LIST all of the different ways that this is possible. ABC ABD…

  10. A B C ABC D ABD C B ACB D ACD D A BDA C BDC • B A C BAC D BCD C A BCA D BCD D A BDA C BDC • C A B CAB D CAD B A CBA D CBD A B DAB C DAC • D A B DAB C DAC B A DBA C DBC C A DCA B DCB 4*3*2 = 24 outcomes Or 4P3 = 24

  11. Combination Examples 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and 2 will be selected to attend the national math conference. LIST all of the different ways that this is possible. 2. From these 4 people (Anne, Bob, Cindy, Dave), and 3 will be selected to attend the national math conference. LIST all of the different ways that this is possible.

  12. Combination answers 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and 2 will be selected to attend the national math conference. LIST all of the different ways that this is possible. AB AC BC AD BD CD 4C2= 6

  13. Combination answer 2. From these 4 people (Anne, Bob, Cindy, Dave), and 3 will be selected to attend the national math conference. LIST all of the different ways that this is possible. ABC BCD ABD ACD 4C3 = 4

  14. Permutations and Combinations • Permutations • Use when ORDER matters and NO repitition • nPr = n!/(n-r)! • Example: If 10 people join a club, how many ways could we pick pres and vp? 10P2 = 90 • Combinations • Use: ORDER does NOT matter and NO repitition • nCr = n!/ [(n-r)!r!] • Example: 10 people join a club. In how many ways could we pick 2? 10C2 = 45

  15. Multiplication, Permutation, or Combination? 1. With 14 players on a team, how many ways could we pick a batting order of 11? 2.      If license plates have 3 letters and then 4 numbers, how many different license plates exist? 3.      How many different four-letter radio station call letters can be formed if the first letter must be W or K? 4.      A social security number contains nine digits.  How many different ones can be formed? 5.      If you wish to arrange your 7 favorite books on a shelf, how many different ways can this be done?

  16. 6.   If you have 10 favorite books, but only have room for 7 books on the shelf, how many ways can you arrange them?   7.      You wish to arrange 12 of your favorite photographs on a mantel.  How many ways can this be done?   8.      You have 20 favorite photographs and wish to arrange 12 of them on a mantel.  How many ways can that be done?  9.  You take a multiple choice test with 12 questions (and each can be answered A B C D E).  How many different ways could you answer the test? 10. If you had 13 pizza toppings, how many ways could you pick 5 of them?

  17. Answers • 14P11   6. 10P7 2.   26*26*26*10*10*10*10 7. 12! or 12P12 3.    2*26*26*26 8.      20P12 4. 10^9   9. 5^12 5. 7! Or 7P7 10. 13 C5  

  18. Review- which method do we use with … • Order matters, repetition allowed • Order matters, repetition NOT allowed • Order DOESN’T matter, repetition allowed • Order DOESN’T matter, repetition NOT allowed

  19. Answers • Order matters, repetition allowed • Multiplication Rule • Ex: Social Security numbers • Order matters, repetition NOT allowed • Permutations: P(n,r)= n!/(n-r)! • Ex: number of ways to pick 1st, 2nd, 3rd from 30 • Order DOESN’T matter, repetition allowed • ??? See section 5.5 • Order DOESN’T matter, repetition NOT allowed • Combinations: C(n,r)= n!/ [(n-r)!*r!] • Ex: number of ways to pick a committee of 3 from 30

  20. Harder problems- if time 1. How many bit strings (0s and 1s) are there are length 5? 2. How many license plates exist if repetition is allowed and they can be 3 letters, 3 digits OR 3 letters, 4 digits? 3. In how many ways can a photographer arrange 8 people in a row from a group of 12? If the bride must be in the picture? If both the bride and groom must be in the picture?

  21. More examples 4. How many passwords exist with 6 letters (and no restrictions)? 5. How many 6 letter passwords exist if: There is no repetition allowed? All are consonants, and repetition is allowed? There is exactly one vowel? There is at least one vowel? (use the COMPLEMENT RULE here) 6. How many vanity plates are there with 4 to 7 characters (letters or digits) with at least 1 digit

  22. answer 1. 2^5 2. 26^3*10^3 + 26^3*10^4 =193,336,000 3. 12*11*10*9*8*7*6*5=19,958,400 8*11P7=8*1,663,200=13,305,600 8*7*10P6=8,467,200

  23. 4. 26^6 26P6 21^6 6*5*21^5 26^6-21^6 Review of complement rule: |E|=total = |E complement| 5. uses product, sum, and complement rule (36^4-26^4)+(36^5-26^5)+(36^6-26^6)+ (36^7-26^7) = 550,466,822,600

  24. More complicated 5.3 problems • Next class

More Related