1. Warm up How many ways can 8 children be placed on a 8- horse Merry-Go-Round? 7! = 5 040 What if Simone insisted on riding the red horse? Here we are only arranging 7 children on 7 horses, so 6! = 720

2. Counting Techniques and Probability Strategies - Combinations Chapter 4.7 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U

3. When Order is Not Important A combination is an unordered selection of elements from a set There are many times when order is not important Suppose Mr. Russell has 10 basketball players and must choose a starting lineup of 5 players (without specifying positions) Order of players is not important We use the notation C(n,r) or n C r where n is the number of elements in the set and r is the number we are choosing

4. Combinations A combination of 5 players from 10 is calculated the following way, giving 252 ways for Mr. Russell to choose his starting lineup

5. An Example of a Restriction on a Combination Suppose that one of Mr. Russell’s players is the superintendent’s daughter, and so must be one of the 5 starting players Here there are really only 4 choices from 9 players So the calculation is C(9,4) = 126 Now there are 126 possible combinations for the starting lineup

6. Combinations from Complex Sets If you can choose of 1 of 3 entrees, 3 of 6 vegetables and 2 of 4 desserts for a meal, how many possible combinations are there? Combinations of entrees = C(3,1) = 3 Combinations of vegetables = C(6,3) = 20 Combinations of desserts = C(4,2) = 6 Possible combinations =  C(3,1) x C(6,3) x C(4,2) = 3 x 20 x 6 = 360 You have 360 possible dinner combinations, so you had better get eating!

7. Calculating the Number of Combinations Suppose you are playing coed volleyball, with a team of 4 men and 5 women The rules state that you must have at least 3 women on the floor at all times (6 players) How many combinations of team lineups are there? You need to take into account team combinations with 3, 4, or 5 women

8. Solution 1: Direct Reasoning In direct reasoning, you determine the number of possible combinations of suitable outcomes and add them Find the combinations that have 3, 4 and 5 women and add them

9. Solution 2: Indirect Reasoning In indirect reasoning, you determine the total possible combinations of outcomes and subtract unsuitable combinations Find the total combinations and subtract those with 2 women

10. Finding Probabilities Using Combinations What is the probability of drawing a Royal Flush (10-J-Q-K-A from the same suit) from a deck of cards? There are C(52,5) ways to draw 5 cards There are 4 ways to draw a royal flush P(Royal Flush) = 4 / C(52,5) = 1 / 649 740 You will likely need to play a lot of poker to get one of these hands!

11. Finding Probability Using Combinations What is the probability of drawing 4 of a kind? There are 13 different cards that can be used to make up the 4 of a kind, and the last card can be any other card remaining

12. Probability and Odds These two terms have different uses in math Probability involves comparing the number of favorable outcomes with the total number of possible outcomes If you have 5 green socks and 8 blue socks in a drawer the probability of drawing a green sock is 5/13 Odds compare the number of favorable outcomes with the number of unfavorable With 5 green and 8 blue socks, the odds of drawing a green sock is 5 to 8 (or 5:8)

13. Combinatorics Summary In Permutations, order matters e.g., President In Combinations, order doesn’t matter e.g., Committee

14. 4 3 2 1 12345 How many routes are there to the top right-hand corner?

15. Permutation AND Combination?? As a permutation:  There are 9 moves  9! However, 4 are identical Up moves and 5 are identical Across moves  Divide by 4! due to identical arrangements of Up  Divide by 5! due to identical arrangements of Across   9! ÷ 4!5! = 126 As a combination:  There are 9 moves required _ _ _ _ _ _ _ _ _  Choose 4 to be Up, the rest are Across  C(9, 4) = 126  Choose 5 to be Across, the rest are Up  C(9, 5) = 126

16. Therefore you can use a permutations and or combinations to solve counting problems.

17. Homework p. 262 – 265 # 1, 2, 3, 5, 7, 9, 18, 23 1 abc, 3, 4, 6a,8

18. References –The Birthday Paradox Wikipedia (2004). Online Encyclopedia. Retrieved September 1, 2004 from http://en.wikipedia.org/wiki/Main_Page

19. The Birthday Paradox I will bet any takers $1 that two people in the room have the same birthday (I didn’t check beforehand). In the long run, will I win money? lose money? break even?

20. The Birthday Paradox Assume all birthdays are equally likely Find the probability that no two birthdays are the same: because the second person cannot have the same birthday as the first (364/365), the third cannot have the same birthday as the first two (363/365), etc.

21. The Birthday Paradox The event of at least two of the n persons having the same birthday is complementary to all n birthdays being different. Therefore, its probability p(n) is The approximate probability that no two people share a birthday in a group of n people. This probability surpasses 1/2 for n = 23 (with value about 50.7%).

22. The Birthday Paradox np(n) 1011.7% 2041.1% 2350.7% 5070.6% 5397.0% 5799.0% 10099.99997% 366100%