1. Counting Principles

2. I. Basic Counting Problem  Say that I have a jar with 7 balls in it numbered 1 through 7. How many ways can 10 be made (by addition of the numbers on the balls) if I pull a ball out of the jar, put it back, and then draw out a second ball.

3. II. Fundamental Principle of Counting  Definition: Let A and B be two events. Say event A can occur in m different ways. After A has occurred, say event B can occur in n different ways. Then the number of ways that the two events can occur is m·n.

4. II. Fundamental Principle of Counting  EXAMPLE: (like #9,11 on HW)  You want to redesign your Facebook profile and change the layout to see only 3 applications. You want one application that pertains to your friends, one that pertains to music, and one that pertains to pictures/videos.  If these are the applications you have available in each category, how many layouts are possible:  Music: 1. iLike 2. My Band 3. Profile Song  Friends: 1. Top Friends 2. Compare Friends  Pictures/Videos: 1. My Flickr 2. Slideshows 3. Graffiti 4. Animoto Videos  Answer: 3x2x4=24

5. II. Fundamental Principle of Counting  EXAMPLE: (like #19 on HW)  Your little brother just broke into your Facebook account and now you need to change your password. You want it to be 5 characters long. The first 3 characters are to be letters and the last 2 characters are to be numbers. How many passwords are possible? What if you don’t want to repeat letters and numbers?  Answer: 26x26x26x10x10=1,757,600  Answer: 26x25x24x10x9=1,404,000

6. II. Fundamental Principle of Counting  EXAMPLE:  Say that you have 42 friends on Facebook. Any time you access your profile, 6 of your friends will appear in a box below your picture. Each time these friends are randomly generated out of your 42 friends. How many different arrangements of 6 people can show up?  Answer:  42x41x40x39x38x37= 3,776,965,920

7. Just to note…  This can also be thought of as: n P r  Permutations of n objects taken r at a time.  For this example we would have 42 P 6 which would get you the same answer

8. II. Fundamental Principle of Counting  EXAMPLE: (like #15,19 on HW)  What if your friend Ally or boyfriend Bobby must be in position one, that is, they are your only 2 choices for that first spot? (note: you still have 42 friends)  Answer:  2x41x40x39x38x37= 179,855,520

9. III. Permutations  Aka: Arranged in order  This is determining the number of ways n elements can be arranged in order.  Definition of Permutation: An ordering of n different elements such that one element if 1 st, one is 2 nd, one is 3 rd and so on.

10. III. Permutations  EXAMPLE: (like #23 (a) on HW)  You want to upload these 6 pictures on Facebook. How many permutations are possible for the 6 pictures?  Answer: 6x5x4x3x2x1=6!=720

11. III. Permutations  So….. the number of permutations of n elements is given by: n·(n-1)……4·3·2·1=n!

12. III. Permutations  EXAMPLE: (like #23 (b) on HW)  You want to upload these 6 pictures on Facebook, but this time you want the two dog pictures have to be next to each other, the two cat pictures have to be next to each other, and the two scenic pictures have to be next to each other? How many permutations are possible for the 6 pictures?  Answer: 3!x2x2x2=48

13. IV. Distinguishable Permutations BANANA BANANAABNANA (if I switch 1 st and 2 nd letters) What happens if I now switch the first and last letters of ABNANA?

14. IV. Distinguishable Permutations  The number of distinguishable permutations of n objects is:  Where n 1, n 2, …, n k are the amounts of the numbers or letters that are repeated.  For BANANA we have 6!/(3!2!)=60 distinct permutations, because we have 2 N’s and 3 A’s in BANANA and 6 letter total.

15. IV. Distinguishable Permutations  EXAMPLE: (like #47, 49 on HW)  So your mom just got Facebook…not cool! You want to disguise your name by scrambling up the letters in you first and last name so she can’t find you. If your name is, Hannah Christopherson, how many new first and last names can you make with the letters?  Hannah: 6!/(2!2!2!)=90  Christopherson: 14!/(2!2!2!2!)=5,448,643,2000

16. V. Combinations  Order is not important (permutations it is) {A, B, C} v. {B, A, C} you would only choose one since both have the same three letters.  Combinations of n elements taken r at a time is: n C r =n!/(n-r)!r!  So you are collecting subsets of a larger set

17. V. Combinations  EXAMPLE: You are having a party this weekend and you want to invite 25 of your 30 Facebook friends. How many groups of 25 can you choose from your 30 friends?  Answer 30 C 25 =142,506

18. Counting in our life…  Cards  Number of possible poker hands….  Where order matters?  Where order doesn’t matter?  Can use counting to find the probability of a certain poker hand, like a full house or a flush.  Combination Locks  Say you have a 39 number combination lock on your bike. What is the probability someone could steal your bike by figuring out your code? License plates, zip codes, phone numbers, etc….. Say a zip code must fit this criteria: 1 st digit: 2-9 3rd digit: 0-9 2 nd digit: 0-9 4th digit: 4-9 5th digit: 1-8