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! =Calculator: #, MATH, PRB, 4:!, ENTERFind 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)?

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Examples: Find the following :• 1. 4! = 4·3·2·1 = 24• 2. 6! =Calculator: #, MATH, PRB, 4:!, ENTERFind 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)?

.

Examples: Find the following :• 1. 4! = 4·3·2·1 = 24• 2. 6! = 6·5·4·3·2·1 = 720Calculator: #, MATH, PRB, 4:!, ENTERFind 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)?

.

Examples: Find the following :• 1. 4! = 4·3·2·1 = 24• 2. 6! = 6·5·4·3·2·1 = 720Calculator: #, MATH, PRB, 4:!, ENTERFind 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)?

.

Examples: Find the following :• 1. 4! = 4·3·2·1 = 24• 2. 6! = 6·5·4·3·2·1 = 720Calculator: #, MATH, PRB, 4:!, ENTERFind 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)?

.

Examples: Find the following :• 1. 4! = 4·3·2·1 = 24• 2. 6! = 6·5·4·3·2·1 = 720Calculator: #, MATH, PRB, 4:!, ENTERFind 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)?.

Examples: Find the following :• 1. 4! = 4·3·2·1 = 24• 2. 6! = 6·5·4·3·2·1 = 720Calculator: #, MATH, PRB, 4:!, ENTERFind 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

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

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 time₁₀P₃ = 10!/(10-3!) = 10!/7! =

• Calc: n, MATH, → PRB, 2:nPr, r, ENTER

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 time₁₀P₃ = 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

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 time₁₀P₃ = 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, ENTER10, MATH, → PRB, 2:nPr, 3, ENTER = 720

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?

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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?

.

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?

.

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?

.

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?

.

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 =

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

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?

.

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!)

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

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?

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)

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!)

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