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3.4B-Permutations

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. Examples: Find the following :. 1. 4! = 2. 6! =

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3.4B-Permutations

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  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 1st, 2nd & 3rd? • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? • 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 1st, 2nd & 3rd? ₈P₃ = • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? • 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 1st, 2nd & 3rd? ₈P₃ = 336 • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? • 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 1st, 2nd & 3rd? ₈P₃ = 336 • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? 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 1st, 2nd & 3rd? ₈P₃ = 336 • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? 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 1st, 2nd & 3rd? ₈P₃ = 336 • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? 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 1st, 2nd & 3rd? ₈P₃ = 336 • 2. If 43 cars start the Daytona 500, how many ways can 3 finish 1st, 2nd & 3rd? 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

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