1. Warm Up Find the number of possible outcomes. 1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna 2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham 3. In how many ways you can choose 4 letters for a password if the first letter must be “S”? 16 17,576 12 Course 3 Permutations and Combinations

2. Course 3 Permutations The factorial of a number is the product of all the whole numbers from the number down to 1. The factorial of 0 is defined to be 1. 5!54321 5! = 5 4 3 2 1 Read 5! as “five factorial.”

3. Evaluate each expression. Example A & B: Evaluating Expressions Containing Factorials Course 3 Permutations A. 8! 8 7 6 5 4 3 2 1 = 40,320 8! 6! 8 7 6 5 4 3 2 1 6 5 4 3 2 1 Write out each factorial and simplify. 8 7 = 56 B. Multiply remaining factors.

4. Example C: Evaluating Expressions Containing Factorials Course 3 Permutations Subtract within parentheses. 10 9 8 = 720 10! 7! 10 9 8 7 6 5 4 3 2 1 7  6  5  4  3  2  1 C. 10! (9 – 2)!

5. Evaluate each expression. Try This: Example 1A & 1B Course 3 Permutations A. 10! 10 9 8 7 6 5 4 3 2 1 = 3,628,800 7! 5! 7 6 5 4 3 2 1 5 4 3 2 1 Write out each factorial and simplify. 7 6 = 42 B. Multiply remaining factors.

6. Try This: Example 1C Course 3 Permutations Subtract within parentheses. 9 8 7 = 504 9! 6! 9 8 7 6 5 4 3 2 1 6  5  4  3  2  1 C. 9! (8 – 2)!

7. Course 3 Permutations A permutation is an arrangement of things in a certain order. If no letter can be used more than once, there are 6 permutations of the first 3 letters of the alphabet: ABC, ACB, BAC, BCA, CAB, and CBA. first letter ? second letter ? third letter ? 3 choices2 choices1 choice The product can be written as a factorial. 3 2 1 = 3! = 6

8. Course 3 Permutations If no letter can be used more than once, there are 60 permutations of the first 5 letters of the alphabet, when taken 3 at a time: ABE, ACD, ACE, ADB, ADC, ADE, and so on. first letter ? second letter ? third letter ? 5 choices4 choices3 choices Notice that the product can be written as a quotient of factorials.  60 = 5 4 3 = = 60 permutations 5 4 3 2 1 2 1 = 5! 2!

9. Course 3 Permutations

10. Jim has 6 different books. Additional Example 2A: Finding Permutations Course 3 Permutations A. Find the number of orders in which the 6 books can be arranged on a shelf. 720 6! (6 – 6)! = 6! 0! = 6 5 4 3 2 1 1 = 6 P 6 = The number of books is 6. The books are arranged 6 at a time. There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

11. Additional Example 2B: Finding Permutations Course 3 Permutations B. If the shelf has room for only 3 of the books, find the number of ways 3 of the 6 books can be arranged. There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways. 6 5 4 6! (6 – 3)! = 6! 3! = 6 5 4 3 2 1 3 2 1 = 6 P 3 = The number of books is 6. The books are arranged 3 at a time. = 120

12. Course 3 Permutations = 5040 7! (7 – 7)! = 7! 0! = 7 6 5 4 3 2 1 1 7 P 7 = The number of cans is 7. The cans are arranged 7 at a time. There are 5040 orders in which to arrange 7 soup cans. Try This: Example 2A A. Find the number of orders in which all 7 soup cans can be arranged on a shelf. There are 7 soup cans in the pantry.

13. Course 3 Permutations There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways. = 7 6 5 4 7! (7 – 4)! = 7! 3! = 7 6 5 4 3 2 1 3 2 1 7 P 4 = The number of cans is 7. The cans are arranged 4 at a time. = 840 There are 7 soup cans in the pantry. Try This: Example 2B B. If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.

14. Course 3 Combinations A combination is a selection of things in any order.

15. Course 3 Combinations If no letter is used more than once, there is only 1 combination of the first 3 letters of the alphabet. ABC, ACB, BAC, BCA, CAB, and CBA are considered to be the same combination of A, B, and C because the order does not matter. If no letter is used more than once, there are 10 combinations of the first 5 letters of the alphabet, when taken 3 at a time. To see this, look at the list of permutations on the next slide.

16. Course 3 Combinations ABCABDABEACDACEADEBCDBCEBDECDE ACBADBAEBADCAECAEDBDCBECBEDCED BACBADBAECADCAEDAECBDCBEDBEDCE BCABDABEACDACEADEADBCCEBDEBDEC CABDABEABDACEACEADDCBEBCEBDECD CBADBAEBADCAECAEDADBCECBEDBEDC These 6 permutations are all the same combination. In the list of 60 permutations, each combination is repeated 6 times. The number of combinations is = 10. 60 6

17. Course 3 Combinations

18. Additional Example 3A: Finding Combinations Course 3 Combinations Mary wants to join a book club that offers a choice of 10 new books each month. A. If Mary wants to buy 2 books, find the number of different pairs she can buy. 10 possible books 2 books chosen at a time 10! 2!(10 – 2)! = 10! 2!8! = 10 9 8 7 6 5 4 3 2 1 (2 1)(8 7 6 5 4 3 2 1) 10 C 2 = = 45 There are 45 combinations. This means that Mary can buy 45 different pairs of books.

19. Additional Example 3B: Finding Combinations Course 3 Combinations B. If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy. 10 possible books 7 books chosen at a time 10! 7!(10 – 7)! = 10! 7!3! 10 C 7 = 10 9 8 7 6 5 4 3 2 1 (7 6 5 4 3 2 1)(3 2 1) = = 120 There are 120 combinations. This means that Mary can buy 120 different sets of 7 books.

20. Try This: Example 3A Course 3 Combinations Harry wants to join a DVD club that offers a choice of 12 new DVDs each month. A. If Harry wants to buy 4 DVDs, find the number of different sets he can buy. 12 possible DVDs 4 DVDs chosen at a time 12! 4!(12 – 4)! = 12! 4!8! = 12 11 10 9 8 7 6 5 4 3 2 1 (4 3 2 1)(8 7 6 5 4 3 2 1) 12 C 4 = = 495

21. Try This: Example 3A Continued Course 3 Combinations There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

22. Try This: Example 3B Course 3 Combinations B. If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy. 12 possible DVDs 11 DVDs chosen at a time 12! 11!(12 – 11)! = 12! 11!1! = 12 11 10 9 8 7 6 5 4 3 2 1 (11 10 9 8 7 6 5 4 3 2 1)(1) 12 C 11 = = 12

23. Try This: Example 3B Continued Course 3 Combinations There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

24. Evaluate each expression. 1. 9! 2. 3. There are 8 hot air balloons in a race. In how many possible orders can all 8 hot air balloons finish the race? 4. A group of 12 people are forming a committee. How many different 4-person committees can be formed? Lesson Quiz 3024 362,880 Insert Lesson Title Here 40,320 495 Course 3 Combinations 9! 5!