1. Counting Fundamentals Ginger Holmes Rowell, Middle TN State University MSP Workshop June 2006

2. Objectives Solve basic problems that involve counting techniques, including the multiplication principle, permutations, and combinations; use counting techniques to understand various situations (e.g., number of ways to order a set of objects, to choose a subcommittee from a committee, to visit n cities)Solve basic problems that involve counting techniques, including the multiplication principle, permutations, and combinations; use counting techniques to understand various situations (e.g., number of ways to order a set of objects, to choose a subcommittee from a committee, to visit n cities)

3. Overview Fundamentals of CountingFundamentals of Counting Multiplication PrincipalMultiplication Principal Permutations: ordered arrangementsPermutations: ordered arrangements Combinations: unordered arrangementsCombinations: unordered arrangements Selected ActivitiesSelected Activities

4. Counting Techniques Fundamentals of CountingFundamentals of Counting Permutations : ordered arrangementsPermutations : ordered arrangements Combinations : unordered arrangementsCombinations : unordered arrangements

5. Fundamentals of Counting Q: Jill has 9 shirts and 4 pairs of pants. How many different outfits does she have?Q: Jill has 9 shirts and 4 pairs of pants. How many different outfits does she have? A: 9 x 4 = 36A: 9 x 4 = 36 36 different outfits 36 different outfits

6. Fundamentals of Counting Multiplication Principle:Multiplication Principle: If there are a ways of choosing one thing, and b ways of choosing a second thing after the first is chosen, then the total number of choice patterns is: If there are a ways of choosing one thing, and b ways of choosing a second thing after the first is chosen, then the total number of choice patterns is: a x b a x b

7. Fundamentals of Counting Q: 3 freshman, 4 sophomores, 5 juniors, and 2 seniors are running for SGA representative. One individual will be selected from each class. How many different representative orderings are possible?Q: 3 freshman, 4 sophomores, 5 juniors, and 2 seniors are running for SGA representative. One individual will be selected from each class. How many different representative orderings are possible? A: 3 · 4 · 5 · 2 = 120 different representative orderingsA: 3 · 4 · 5 · 2 = 120 different representative orderings

8. Generalized Multiplication Principle If there are a ways of choosing one thing, b ways of choosing a second thing after the first is chosen, and c ways of choosing a third thing after the first two have been chosen…and z ways of choosing the last item after the earlier choices, then the total number of choice patterns isIf there are a ways of choosing one thing, b ways of choosing a second thing after the first is chosen, and c ways of choosing a third thing after the first two have been chosen…and z ways of choosing the last item after the earlier choices, then the total number of choice patterns is a x b x c x … x z

9. Example Q: When I lived in Madison Co., AL, the license plates had 2 fixed numbers, 2 variable letters and 3 variable numbers. How many different license plates were possible?Q: When I lived in Madison Co., AL, the license plates had 2 fixed numbers, 2 variable letters and 3 variable numbers. How many different license plates were possible? A: 26 x 26 x 10 x 10 x 10 = 676,000 different platesA: 26 x 26 x 10 x 10 x 10 = 676,000 different plates

10. Fundamentals of Counting Q: How many more license plate numbers will Madison County gain by changing to 3 letters and 2 numbers? A: 26 x 26 x 26 x 10 x 10 = 1,757,600 1,757,600 – 676,000 = 1,081,600 more license plate numbers 1,757,600 – 676,000 = 1,081,600 more license plate numbers

11. Permutations: Ordered Arrangements Q: Given 6 people and 6 chairs in a line, how many seating arrangements (orderings) are possible?Q: Given 6 people and 6 chairs in a line, how many seating arrangements (orderings) are possible? A: 6 · 5 · 4 · 3 · 2 · 1 = 6! = 720 orderingsA: 6 · 5 · 4 · 3 · 2 · 1 = 6! = 720 orderings

12. Permutations: Ordered Arrangements Q: Given 6 people and 4 chairs in a line, how many different orderings are possible?Q: Given 6 people and 4 chairs in a line, how many different orderings are possible? A: 6 · 5 · 4 · 3 = 360 different orderingsA: 6 · 5 · 4 · 3 = 360 different orderings

13. Permutations: Ordered Arrangements Permutation of n objects taken r at a time:Permutation of n objects taken r at a time: r-permutation, P(n,r), n P r Q: Given 6 people and 5 chairs in a line, how many different orderings are possible? A: 6 · 5 · 4 · 3 · 2 = 720 different orderings

14. Permutations: Ordered Arrangements n P r = n(n-1)···(n-(r-1)) n P r = n(n-1)···(n-(r-1)) = n(n-1)···(n-r+1) = n(n-1)···(n-r+1) = n(n-1)···(n-r+1) (n-r)! = n(n-1)···(n-r+1) (n-r)! (n-r)! (n-r)! = n(n-1)···(n-r+1)(n-r)···(3)(2)(1) = n(n-1)···(n-r+1)(n-r)···(3)(2)(1) (n-r)! (n-r)! = n! = n! (n-r)! (n-r)!

15. Permutations: Ordered Arrangements Q: How many different batting orders are possible for a baseball team consisting of 9 players?Q: How many different batting orders are possible for a baseball team consisting of 9 players? A: 9 · 8 · 7 ··· 3 · 2 · 1 = 9!A: 9 · 8 · 7 ··· 3 · 2 · 1 = 9! = 362,880 batting orders = 362,880 batting orders Note: this is equivalent to 9 P 9.Note: this is equivalent to 9 P 9. 9 P 9 = 9! = 9! = 9! 9 P 9 = 9! = 9! = 9! (9-9)! 0! (9-9)! 0!

16. Permutations: Ordered Arrangements Q: How many different batting orders are possible for the leading four batters?Q: How many different batting orders are possible for the leading four batters? A: 9 · 8 · 7 · 6 = 3,024 ordersA: 9 · 8 · 7 · 6 = 3,024 orders 9 P 4 = 9! = 9! = 9! 9 P 4 = 9! = 9! = 9! (9-4)! 5! 5! (9-4)! 5! 5!

17. Permutations: Indistinguishable Objects Q: How many different letter arrangements can be formed using the letters T E N N E S S E E ?Q: How many different letter arrangements can be formed using the letters T E N N E S S E E ? A: There are 9! Permutations of the letters T E N N E S S E E if the letters are distinguishable.A: There are 9! Permutations of the letters T E N N E S S E E if the letters are distinguishable. However, 4 E’s are indistinguishable. There are 4! ways to order the E’s.However, 4 E’s are indistinguishable. There are 4! ways to order the E’s.

18. Permutations: Indistinguishable Objects, Cont. 2 S’s and 2 N’s are indistinguishable. There are 2! orderings of each.2 S’s and 2 N’s are indistinguishable. There are 2! orderings of each. Once all letters are ordered, there is only one place for the T.Once all letters are ordered, there is only one place for the T. If the E’s, N’s, & S’s are indistinguishable among themselves, then there areIf the E’s, N’s, & S’s are indistinguishable among themselves, then there are 9! = 3,780 different orderings of (4!·2!·2!) T E N N E S S E E 9! = 3,780 different orderings of (4!·2!·2!) T E N N E S S E E

19. Permutations: Indistinguishable Objects Subsets of Indistinguishable Objects Subsets of Indistinguishable Objects Given n objects of which Given n objects of which a are alike, b are alike, …, and a are alike, b are alike, …, and z are alike z are alike There are n! permutations. There are n! permutations. a!·b!···z! a!·b!···z!

20. Combinations: Unordered Arrangements Combinations: number of different groups of size r that can be chosen from a set of n objects (order is irrelevant)Combinations: number of different groups of size r that can be chosen from a set of n objects (order is irrelevant) Q: From a group of 6 people, select 4. How many different possibilities are there?Q: From a group of 6 people, select 4. How many different possibilities are there? A: There are 6 P 4 =360 different orderings of 4 people out of 6.A: There are 6 P 4 =360 different orderings of 4 people out of 6. 6·5·4·3 = 360 = 6 P 4 = 6·5·4·3 = 360 = 6 P 4 = n! (n -r)!

21. Unordered Example continued However the order of the chosen 4 people is irrelevant. There are 24 different orderings of 4 objects.However the order of the chosen 4 people is irrelevant. There are 24 different orderings of 4 objects. 4 · 3 · 2 · 1 = 24 = 4! =r! Divide the total number of orderings by the number of orderings of the 4 chosen people. 360 = 15 different groups of 4 people.Divide the total number of orderings by the number of orderings of the 4 chosen people. 360 = 15 different groups of 4 people. 24 24

22. Combinations: Unordered Arrangements The number of ways to choose r objects from a group of n objects. The number of ways to choose r objects from a group of n objects. C(n,r) or n C r, read as “n choose r” C(n,r) or n C r, read as “n choose r”

23. Combinations: Unordered Arrangements Q: From a group of 20 people, a committee of 3 is to be chosen. How many different committees are possible?Q: From a group of 20 people, a committee of 3 is to be chosen. How many different committees are possible? A:A:

24. Combinations: Unordered Arrangements Q: From a group of 5 men & 7 women, how many different committees of 2 men & 3 women can be found?Q: From a group of 5 men & 7 women, how many different committees of 2 men & 3 women can be found? A: There are 5 C 2 groups of men & 7 C 3 groups of women. Using the multiplication principleA: There are 5 C 2 groups of men & 7 C 3 groups of women. Using the multiplication principle

25. Practice Problem You have 30 students in your class, which will be arranged in 5 rows of 6 people. Assume that any student can sit in any seat.You have 30 students in your class, which will be arranged in 5 rows of 6 people. Assume that any student can sit in any seat. How many different seating charts could you have for the first row?How many different seating charts could you have for the first row? How many different seating charts could you have for the whole class?How many different seating charts could you have for the whole class?

26. It’s Your Turn Make up three counting problems which would interest your students, include one permutation and one combination and one of your choice.Make up three counting problems which would interest your students, include one permutation and one combination and one of your choice. Calculate the answer for these problems.Calculate the answer for these problems.

27. Questions ??? Fundamentals of Counting?Fundamentals of Counting? Permutations?Permutations? Combinations?Combinations?

28. Homework