1. 3.4B-Permutations Permutation: ORDERED arrangement of objects. # of different permutations (ORDERS) of n distinct objects is n! n! = n(n-1)(n-2)(n-3)…3·2·1 0! = 1 (special) 1! = 1 2! = 2·1 3! = 3·2·1 etc.

2. Examples: Find the following : 1. 4! = 2. 6! = Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 4. 10! = 5. How many different batting orders are there for a 9 player team? 6. There are 6 teams in the division. How many different orders can they finish in (no ties)?.

3. Examples: Find the following : 1. 4! = 4·3·2·1 = 24 2. 6! = Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 4. 10! = 5. How many different batting orders are there for a 9 player team? 6. There are 6 teams in the division. How many different orders can they finish in (no ties)?.

4. Examples: Find the following : 1. 4! = 4·3·2·1 = 24 2. 6! = 6·5·4·3·2·1 = 720 Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 4. 10! = 5. How many different batting orders are there for a 9 player team? 6. There are 6 teams in the division. How many different orders can they finish in (no ties)?.

5. Examples: Find the following : 1. 4! = 4·3·2·1 = 24 2. 6! = 6·5·4·3·2·1 = 720 Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 40,320 4. 10! = 5. How many different batting orders are there for a 9 player team? 6. There are 6 teams in the division. How many different orders can they finish in (no ties)?.

6. Examples: Find the following : 1. 4! = 4·3·2·1 = 24 2. 6! = 6·5·4·3·2·1 = 720 Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 40,320 4. 10! = 3,628,800 5. How many different batting orders are there for a 9 player team? 6. There are 6 teams in the division. How many different orders can they finish in (no ties)?.

7. Examples: Find the following : 1. 4! = 4·3·2·1 = 24 2. 6! = 6·5·4·3·2·1 = 720 Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 40,320 4. 10! = 3,628,800 5. How many different batting orders are there for a 9 player team? 9! = 362,880 6. There are 6 teams in the division. How many different orders can they finish in (no ties)?.

8. Examples: Find the following : 1. 4! = 4·3·2·1 = 24 2. 6! = 6·5·4·3·2·1 = 720 Calculator: #, MATH, PRB, 4:!, ENTER Find with the calculator: 3. 8! = 40,320 4. 10! = 3,628,800 5. How many different batting orders are there for a 9 player team? 9! = 362,880 6. There are 6 teams in the division. How many different orders can they finish in (no ties)? 6! = 720

9. Permutation of n objects taken r at a time Number of ways to choose SOME objects in a group and put them in ORDER. nPr = n!/(n-r)! n=# objects in group r=# objects chosen Ex: How many ways can 3 digit codes be formed if no repeats? Calc: n, MATH, → PRB, 2:nPr, r, ENTER

10. Permutation of n objects taken r at a time Number of ways to choose SOME objects in a group and put them in ORDER. nPr = n!/(n-r)! n=# objects in group r=# objects chosen Ex: How many ways can 3 digit codes be formed if no repeats? 10 digits, 3 chosen at a time ₁₀P₃ = 10!/(10-3!) = 10!/7! = Calc: n, MATH, → PRB, 2:nPr, r, ENTER

11. Permutation of n objects taken r at a time Number of ways to choose SOME objects in a group and put them in ORDER. nPr = n!/(n-r)! n=# objects in group r=# objects chosen Ex: How many ways can 3 digit codes be formed if no repeats? 10 digits, 3 chosen at a time ₁₀P₃ = 10!/(10-3!) = 10!/7! = (10·9·8·7·6·5·4·3·2·1)/(7·6·5·4·3·2·1) = 10·9·8 = 720 Calc: n, MATH, → PRB, 2:nPr, r, ENTER

12. Permutation of n objects taken r at a time Number of ways to choose SOME objects in a group and put them in ORDER. nPr = n!/(n-r)! n=# objects in group r=# objects chosen Ex: How many ways can 3 digit codes be formed if no repeats? 10 digits, 3 chosen at a time ₁₀P₃ = 10!/(10-3!) = 10!/7! = (10·9·8·7·6·5·4·3·2·1)/(7·6·5·4·3·2·1) = 10·9·8 = 720 Calc: n, MATH, → PRB, 2:nPr, r, ENTER 10, MATH, → PRB, 2:nPr, 3, ENTER = 720

13. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order?.

14. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₈P₃ = 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order?.

15. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₈P₃ = 336 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order?.

16. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₈P₃ = 336 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₄₃P₃ = 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order?.

17. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₈P₃ = 336 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₄₃P₃ = 74,046 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order?.

18. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₈P₃ = 336 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₄₃P₃ = 74,046 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order? ₁₂P₄ =

19. Examples: Use the calculator 1. In a race with 8 horses, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₈P₃ = 336 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1 st, 2 nd & 3 rd ? ₄₃P₃ = 74,046 3. A company has 12 people on the board. How many ways can Pres., V.P., Sec., & Treas. be filled in this order? ₁₂P₄ = 11,880

20. Distinguishable (different) Permutations # of way to put n objects in ORDER where some of the objects are the SAME. n!/(n₁!·n₂!·n₃! …) n₁, n₂, n₃ are different types of objects EX: A subdivision is planned. There will be 12 homes. (6 1-story, 4 2-story & 2 split level) In how many distinguishable ways can they be arranged?.

21. Distinguishable (different) Permutations # of way to put n objects in ORDER where some of the objects are the SAME. n!/(n₁!·n₂!·n₃! …) n₁, n₂, n₃ are different types of objects EX: A subdivision is planned. There will be 12 homes. (6 1-story, 4 2-story & 2 split level) In how many distinguishable ways can they be arranged? 12 total, 3 groups (6,4,2).12!/(6!4!2!)

22. Distinguishable (different) Permutations # of way to put n objects in ORDER where some of the objects are the SAME. n!/(n₁!·n₂!·n₃! …) n₁, n₂, n₃ are different types of objects EX: A subdivision is planned. There will be 12 homes. (6 1-story, 4 2-story & 2 split level) In how many distinguishable ways can they be arranged? 12 total, 3 groups (6,4,2).12!/(6!4!2!) = 13,860

23. Examples: Ex: Trees will be planted in the subdivision evenly spaced. (6 oaks, 9 maples, & 5 poplars) How many distinguishable ways can they be planted?

24. Examples: Ex: Trees will be planted in the subdivision evenly spaced. (6 oaks, 9 maples, & 5 poplars) How many distinguishable ways can they be planted? 20 total, 3 groups (6,9,5)

25. Examples: Ex: Trees will be planted in the subdivision evenly spaced. (6 oaks, 9 maples, & 5 poplars) How many distinguishable ways can they be planted? 20 total, 3 groups (6,9,5) 20!/(6!9!5!)

26. Examples: Ex: Trees will be planted in the subdivision evenly spaced. (6 oaks, 9 maples, & 5 poplars) How many distinguishable ways can they be planted? 20 total, 3 groups (6,9,5) 20!/(6!9!5!) = 77,597,520