2. Aim: Combinations Course: Alg. 2 & Trig. Do Now: Aim: How do we determine the number of outcomes when order is not an issue? Ann, Barbara, Carol, and Dave are the only members of a school club. In how many different ways can they elect a president and treasurer for the club? Ann, Barbara, Carol, and Dave are the only members of a school club. In how many ways can they choose 2 people to represent the club at student council meetings? Explain how these situations are different.

3. Aim: Combinations Course: Alg. 2 & Trig. Permutation Ann, Barbara, Carol, and Dave are the only members of a school club. In how many different ways can they elect a president and treasurer for the club? Ann Barbara Carol Dave President Barbara Carol Dave Ann Carol Dave Barbara Ann Dave Barbara Carol Ann Treasurer. Ann & Barbara Ann & Carol Ann & Dave Barbara & Ann Barbara & Carol Barbara & Dave Carol & Barbara Carol & Ann Carol & Dave Dave & Barbara Dave & Carol Dave & Ann 4 P 2 = 4 3 = 12 There are 12 different arrangements of two people for president and treasurer.

4. Aim: Combinations Course: Alg. 2 & Trig. Combination Ann, Barbara, Carol, and Dave are the only members of a school club. In how many ways can they choose 2 people to represent the club at student council meetings? Ann Barbara Carol Dave 1st Person Barbara Carol Dave Ann Carol Dave Barbara Ann Dave Barbara Carol Ann 2nd Person Ann & Barbara Ann & Carol Ann & Dave Barbara & Ann Barbara & Carol Barbara & Dave Carol & Barbara Carol & Ann Carol & Dave Dave & Barbara Dave & Carol Dave & Ann There are six combinations of two people that can represent

5. Aim: Combinations Course: Alg. 2 & Trig. Order: Permutation vs. Combination A selection of objects in which their order is not important. When selecting some of the objects in the set: The number of combinations of n objects r at a time When selecting all objects in the set: there is only 1 combination!! = 1

6. Aim: Combinations Course: Alg. 2 & Trig. Combinations 1. For any counting number n, n C n = 1 3 C 3 = 1 10 C 10 = 1 2. For any counting number n, n C 0 = 1 5 C 0 = 1 34 C 0 = 1 Some Special Relationships 3. For whole numbers n and r, where r < n, n C r = n C n - r 7 C 3 = 7 C 7 - 3 = 7 C 4 23 C 16 = 23 C 23 - 16 = 23 C 7

7. Aim: Combinations Course: Alg. 2 & Trig. Probability Notation Ann, Barbara, Carol, and Dave are the only members of a school club. In how many ways can they choose 2 people to represent the club at student council meetings? Ann Barbara Carol Dave 1st Person Barbara Carol Dave Ann Carol Dave Barbara Ann Dave Barbara Carol Ann 2nd Person Ann & Dave Barbara & Ann Barbara & Carol Barbara & Dave Carol & Barbara Carol & Ann Carol & Dave Dave & Barbara Dave & Carol Dave & Ann 4C24C2 = 4 P 2 / 2! = 6 Ann & Barbara Ann & Carol

8. Aim: Combinations Course: Alg. 2 & Trig. Model Problems Evaluate: 10 C 38C28C2 How many different three-person committees can be formed from a group of eight people? Is order important? NO 8C38C3 = 56 = 120 = 28 A committee has 7 men and 5 women. A subcommittee of 8 is to be formed. Write an expression for the number of ways the choice can be made. 12 C 8 = 495 In general, use permutations where order is important, and combinations where order is not important.

9. Aim: Combinations Course: Alg. 2 & Trig. Model Problem Is the order of the 4 marbles important? NO! Combination 457 = 2380 From an urn containing 4 black marbles, 8 blue marbles, and 5 red marbles, in how many ways can a set of 4 marbles be selected? 17 C 4 = 17 total marbles

10. Aim: Combinations Course: Alg. 2 & Trig. Model Problem If n C 2 = 15, what is the value of n? nC2nC2 n(n - 1) = 215 6C26C2 = 15 n 2 - n = 30 n 2 - n - 30 = 0 (n - 6)(n + 5) = 0 (n - 6) = 0 (n + 5) = 0 n = 6 n = -5

11. Aim: Combinations Course: Alg. 2 & Trig. Model Problems There are 10 boys and 20 girls in a class. Find the number of ways a team of 3 students can be selected to work on a project if the team consists of: 30 C 3 10 C 120 C 2 20 C 3 + 10 C 120 C 2 = 4060 = 10 190 = 1900 = 1140 10 C 0 = 1900 + 1140 = 3040 A. Any 3 students B. 1 boy and 2 girls C. 3 girls D. At least 2 girls 2 girls 10 C 0 3 girls

12. Aim: Combinations Course: Alg. 2 & Trig. Model Problem In how many ways can 6 marbles be distributed in 3 boxes so that 3 marbles are in the first box, 2 in the second, and 1 in the third Box 2Box 1Box 3 6C36C33C23C21C11C1 2031 = 60

13. Aim: Combinations Course: Alg. 2 & Trig. Model Problems In the “Pick Four” Lottery, you create a 4-digit number using the numbers 1, 2, 3, 4, 5, and 6. If you play the game “straight”, you win if the winning lottery number matches your selection exactly. How many different arrangements are possible if you bet the game “straight”? 6P46P4 = 360 If you choose you may, you may play the game “boxed”. This means that as long as the same four numbers are chosen, regardless of order, you win. How many possible combinations are possible? 6C46C4 = 15

14. Aim: Combinations Course: Alg. 2 & Trig. Model Problem How many different 4-member committees can be formed from a group of 10 people if Tony, 1 of the 10 must: A. Always on the committee AFTER TONY IS PLACED ON THE COMMITTEE, THERE ARE 3 PLACES LEFT FOR THE OTHER 9 PEOPLE 1 = 84 9C39C3 9 C 3 = 84 TONY IS A MUST!

15. Aim: Combinations Course: Alg. 2 & Trig. Model Problem How many different 4-member committees can be formed from a group of 10 people if Tony, 1 of the 10 must: B. Never be on the committee 9C49C4 There are now only 9 possible members for the 4-member committee = 126

16. Aim: Combinations Course: Alg. 2 & Trig. Regents Problem The principal would like to assemble a committee of 8 students from the 15-member student council. How many different committees can be chosen? (1) 120 (3) 32,432,400 (2) 6,435 (4) 259,459,200

17. Aim: Combinations Course: Alg. 2 & Trig. Model Problem Find the number of ways to select 5-card hands from a standard deck so that each hand contains at most 2 aces. at most 2 aces Means that the hand could have 0, 1 or 2 aces W/ no Aces 4C04C048 C 5 W/ 1 Aces 4C14C148 C 4 W/ 2 Aces 4C24C248 C 3 Choose AcesComplete the 5-card hand = 1712304 = 778320 = 103776 = 2594400 +

18. Aim: Combinations Course: Alg. 2 & Trig. Probability Involving Combinations Dependent Events 2 Permutations 1 Counting Principle Two cards are drawn at random from a standard deck of 52 cards, without replacement. What is the probability that both cards drawn are fives? P(1st 5) = 4/52P(2d 5) = 3/51 P(5, 5) = 4/52 3/51 Number of ways to draw 2 5’s from possible 4 Number of ways to draw any 2 cards from deck of 52 4C24C252 C 2 4P24P2 52 P 2 = 1/221 3 Combinations 4C24C2 52 C 2 = 1/221

19. Aim: Combinations Course: Alg. 2 & Trig. Model Problem In a school organization, there are 4 sophomores and 5 juniors. A committee of 4 people is to be selected from this group. What is the probability that 2 sophomores and 2 juniors will be on the committee? P(2 sophomores, 2 juniors on the 4-member committee) = n(4-member combinations made from of the 9 members) n(combinations of 2s & 2j) Permutation or Combination

20. Aim: Combinations Course: Alg. 2 & Trig. Model Problem Successful Outcomes n(E): Total Outcomes n(S): Nine total students from which to choose In a school organization, there are 4 sophomores and 5 juniors. A committee of 4 people is to be selected from this group. What is the probability that 2 sophomores and 2 juniors will be on the committee? : 9 C 4 = 126 How do we determine n(2 soph., 2 jun.)? Two of four sophomores 4C24C2 = 6 Two of five juniors 5C25C2 = 10 n(2 soph., 2 jun.) = 4 C 2 5 C 2 = 610 = 60

21. Aim: Combinations Course: Alg. 2 & Trig. Model Problem P(2 sophs., 2 juniors) = In a school organization, there are 4 sophomores and 5 juniors. A committee of 4 people is to be selected from this group. What is the probability that 2 sophomores and 2 juniors will be on the committee? 9 C 4 4 C 2 5 C 2 = 6 10 126 = 60 126 = 10 21

22. Aim: Combinations Course: Alg. 2 & Trig. Model Problem Total Outcomes n(S) : P(at least 2 b) = 5 C 2 An urn contains 4 white marbles and 5 blue marbles, all of equal size. Three marbles are drawn at random with no replacement. What is the probability that at least 2 marbles drawn are blue? 9 C 3 At least 2 are blue: (2-b, 1-w) or (3-b) Successful Outcomes n(E) 5 C 2 4 C 1 + 5C35C3 4 C 1 5 C 3 9 C 3 Total Outcomes n(S) = 84 40 + 10 = 25/42 Successful Outcomes n(E) P(A  B) = P(A) + P(B) - P(A  B)

23. Aim: Combinations Course: Alg. 2 & Trig. Model Problem A candy dish contains 10 candies. Three candies are covered with red foil and 7 with green foil. A. If 2 candies are chosen at random from the dish, what is the probability that both will be covered with the same colored foil? P(both red or both green) = P(2R) + P(2G) P(2R) = 3 C 2 P(2G) = 7 C 2 10 C 2 P(A  B) = P(A) + P(B) - P(A  B) P(2R or 2G) = = 8/15 = 3 + 21 45 10 C 2 3 C 2 + 7 C 2

24. Aim: Combinations Course: Alg. 2 & Trig. Model Problem A candy dish contains 10 candies. Three candies are covered with red foil and 7 with green foil. B. If 2 candies are chosen at random from the dish, what is the probability that each will be covered with a different foil?

25. Aim: Combinations Course: Alg. 2 & Trig. Model Problem A candy dish contains 10 candies. Three candies are covered with red foil and 7 with green foil. C. If 5 candies are chosen at random from the dish, what is the probability that at least 3 will be covered with green foil?