Section 11.1

2. Math in Our World Section 11.1 The Fundamental Counting Principle and Permutations

3. Learning Objectives • Use the fundamental counting principle. • Calculate the value of factorial expressions. • Find the number of permutations of n objects. • Find the number of permutations of n objects taken r at a time. • Find the number of permutations when some objects are alike.

4. Fundamental Counting Principle In a sequence of n events in which the first event can occur in k1ways and the second event can occur in k2ways and the third event can occur in k3ways and so on, the total number of ways the sequence can occur is k1 k2 k3 . . . kn

5. EXAMPLE 1 Using the Fundamental Counting Principle There are four blood types: A, B, AB, and O. Blood is also either Rh+ or Rh-. If a local blood bank labels donations according to type, Rh factor, and gender of the donor, how many different ways can a blood sample be labeled?

6. EXAMPLE 1 Using the Fundamental Counting Principle There are four blood types: A, B, AB, and O. Blood is also either Rh+ or Rh-. If a local blood bank labels donations according to type, Rh factor, and gender of the donor, how many different ways can a blood sample be labeled? SOLUTION There are four possibilities for blood type, two for Rh factor, and two for gender of the donor. Using the fundamental counting principle, there are 4 x 2 x 2 = 16 different ways that blood could be labeled.

7. Ex: Permutation: Arrange books

8. EXAMPLE 2 Using the Fundamental Counting Principle w/Repetition (a) The letters A, B, C, D, and E are to be used in a four-letter ID card. How many different cards are possible if letters are allowed to be repeated? (b) How many cards are possible if each letter can only be used once?

9. EXAMPLE 2 Using the Fundamental Counting Principle w/Repetition SOLUTION (a) There are four spaces to fill and five choices for each. The fundamental counting principle gives us 5 x 5 x 5 x 5 = 54 = 625 (b) The first letter can still be chosen in five ways. But with no repetition allowed, there are only four choices for the second letter, three for the third, and two for the last. The number of potential cards is 5 x 4 x 3 x 2 = 120 Hopefully, this ID card is for a pretty small organization.

10. Factorial Notation For any natural number n n! = n(n – 1)(n – 2)(n – 3) . . . 3 x 2 x 1 n! is read as “n factorial.” 0! is defined as 1. The symbol for a factorial is the exclamation mark (!). In general, n! means to multiply the whole numbers from n down to 1. For example, 1! = 1 = 1 3! = 3 x 2 x 1 = 6 2! = 2 x 1 = 2 4! = 4 x 3 x 2 x 1 = 24

11. Factorial Notation Some of the formulas we’ll be working with require division of factorials. This will be simple if we make two key observations: • You can write factorials without writing all of the factors down to 1. For example, 5! = 5 x 4 x 3 x 2 x 1, but we can also write this as 5 x 4!, or 5 x 4 x 3!, etc.

12. EXAMPLE 3 Evaluating Factorial Expressions Evaluate each expression: (a) 8! (b)

13. EXAMPLE 3 Evaluating Factorial Expressions Evaluate each expression: (a) 8! (b) SOLUTION (a) 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 (b) First, write 12! as 12 x 11 x 10!, then note that

14. Permutations An arrangement of n distinct objects in a specific order is called a permutation of the objects. The number of permutations of n distinct objects using all of the objects is n!.

15. EXAMPLE 4 Calculating the Number of Permutations In seven of the 10 years from 1998–2007, the five major league baseball teams in the American League East division finished in the exact same order: New York, Boston, Toronto, Baltimore, Tampa Bay. Just how unusual is this? Find the number of possible finishing orders for these five teams.

16. EXAMPLE 4 Calculating the Number of Permutations SOLUTION This is a permutation problem since we’re deciding on the number of ways to arrange five distinct objects. There are 5! = 120 possible finishing orders.

17. EXAMPLE 5 Solving a Permutation Problem How many different ways can a pledge class with 20 members choose a president, vice president, and Greek Council representative? (No pledge can hold two offices.)

18. EXAMPLE 5 Solving a Permutation Problem How many different ways can a pledge class with 20 members choose a president, vice president, and Greek Council representative? (No pledge can hold two offices.) SOLUTION There are 20 choices for president, 19 remaining candidates for vice president, and 18 members left to choose from for Greek Council rep. So there are 20 x 19 x 18 = 6,840 different ways to assign these three offices.

19. Permutations Permutation of n Objects Taken r at a Time The arrangement of n objects in a specific order using r of those objects is called a permutation of n objects taken r at a time. It is written as nPr, and is calculated using the formula

20. EXAMPLE 6 Solving a Permutation Problem How many five-digit zip codes are there with no repeated digits? SOLUTION This is a permutation problem because five numbers are taken from 10 possible digits, with order important and no repetition. In this case, n = 10 and r = 5.

21. Permutations Permutation Rule When Objects Are Alike The number of permutations of n objects in which k1objects are alike, k2objects are alike, etc. is where k1+ k2+ . . . + kp= n

22. EXAMPLE 7 Solving a Permutation Problem with Like Objects How many different passwords can be made using all of the letters in the word Mississippi ? SOLUTION The letters can be rearranged as M IIII SSSS PP. Then n = 11, k1= 1, k2= 4, k3= 4, and k4= 2. Using our newest formula, there are different passwords.