1. Math Duels Competition of Scholars

2. Rules  The class must be split into 2 groups.  Each group must select a team captain and they must do Rock-Paper-Scissors to decide who is group 1.  Group 1 then chooses one member from each group to compete against each other.  A problem will be on the board and you must race to complete the problem.  The first person to complete the problem will win the point for the time.  There is no cheating: No copying each others’ work, only one person at the board, your team CANNOT help you once you are at the board.

3. Ready? Set? Go!

4. Given 3 people, Bob, Mike and Sue, how many different ways can these three people be arranged where order matters? Question

5. Answer  Let BMS stand for the order of Bob on the left, Mike in the middle and Sue on the right. Since order matters, a different arrangement is BSM. Where Bob is on the left, Sue is in the middle and Mike is on the right. If we find all possible arrangements of Bob, Mike and Sue where order matters, we have the following:  BMS, BSM, MSB, MBS, SMB, SBM  The number of ways to arrange three people three at a time is: 3! = (3)(2)(1) = 6 ways

6. Question Find the number of ways to arrange 4 people in groups of 3 at a time where order matters.

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8. Question Find the number of ways to arrange 6 items in groups of 4 at a time where order matters.

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10. Question Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter.

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12. Question How many ways are there to select a subcommittee of 7 members from among a committee of 17?

13. Answer  Since it does not matter what order the committee members are chosen in, the combination formula is used. Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation. C(17,7)= 19,448

14. Question Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

15. Answer  When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used. C(52,5) = 2,598,960

16. Question There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain three women and one man?

17. Answer  Since no order to the committee is mentioned, a combination instead of a permutation is used. Lets sort out what we have and what we want. Have: 5 women, 6 men. Want: 3 women AND 1 man. The word AND means multiply. Woman and Men C(have, want)*C(have, want) C(have 5 women, want 3 women)*C(have 6 men, want 1 man) C(5,3)*C(6,1) = 60

18. Question There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain at least three women?

19. Answer  Have: 5 women, 6 men. Want: 3 women AND 1 man OR 4 women and 0 men. C(have, want)*C(have, want) + C(have,want)*C(have, want) C(have 5 women, want 3 women)*C(have 6 men, want 1 man) + C(have 5 women,want 4 women)*C(have 6 men, want 0 men) C(5,3)*C(6,1) + C(5,4)*C(6,0) 60 + 5 = 65

20. Question A school has scheduled three volleyball games, two soccer games, and four basketball games. You have a ticket allowing you to attend three of the games. In how many ways can you go to two basketball games and one of the other events?

21. Answer  Since order does not matter it is a combination. The word AND means multiply.  Given 4 basketball, 3 volleyball, 2 soccer.  We want 2 basketball games and 1 other event. There are 5 choices left.  C(n,r)  C(How many do you have, How many do you want)  C(have 4 basketball, want 2 basketball)*C(have 5 choices left, want 1) C(4,2)*C(5,1) (6)(5) = 30

22. Question In a local election, there are seven people running for three positions. The person that has the most votes will be elected to the highest paying position. The person with the second most votes will be elected to the second highest paying position, and likewise for the third place winner. How many different outcomes can this election have?

23. Answer  This committee has an order! The person with the highest paying position has a higher order over the other two members of the committee, so order matters and a permutation is used. There are 7 people and we want groups of 3 at a time where order matters. P(7,3) = 210 Therefore there are 210 different possible ways this committee can be formed. This is the same as a slot problem without repetition. Since there are three seats to be assigned, there are three slots. The first slot has 7 different ways to assign the seat, the second slot has 6 different ways to assign the seat, and the third slot has 5 different ways to assign the seat. Therefore there are 210 different ways to form the committee.

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