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Section 4-6
Permutations of a Set
Essential Question



How do you find the number of permutations of a set?



Where you’ll see this:

  Cooking, travel, music, sports, games
Vocabulary

1. Permutation:



2. Factorial:



3. n Pr :
Vocabulary

1. Permutation: An arrangement of items in a particular order



2. Factorial:



3. n Pr :
Vocabulary

1. Permutation: An arrangement of items in a particular order
                   ORDER IS IMPORTANT!!!

2. Factorial:



3. n Pr :
Vocabulary

1. Permutation: An arrangement of items in a particular order
                   ORDER IS IMPORTANT!!!

2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)...
    (3)(2)(1)


3. n Pr :
Vocabulary

1. Permutation: An arrangement of items in a particular order
                   ORDER IS IMPORTANT!!!

2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)...
    (3)(2)(1)       5! = (5)(4)(3)(2)(1)


3. n Pr :
Vocabulary

1. Permutation: An arrangement of items in a particular order
                   ORDER IS IMPORTANT!!!

2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)...
    (3)(2)(1)       5! = (5)(4)(3)(2)(1) = 120


3. n Pr :
Vocabulary

1. Permutation: An arrangement of items in a particular order
                   ORDER IS IMPORTANT!!!

2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)...
    (3)(2)(1)       5! = (5)(4)(3)(2)(1) = 120


3. n Pr : n permutations taken r at a time; n is the number of different
      items and r is the number of items taken at a time
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?

                            3
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?

                            3      2
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?

                            3      2      1
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?

                            3       2     1

                        (3)(2)(1)
Example 1
How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no
                     repetitions are allowed?

                            3      2      1

                        (3)(2)(1) = 6 numbers
Permutation Formula
Permutation Formula



              n!
    n
      Pr =
           (n − r)!
Permutation Formula



                                   n!
                         n
                           Pr =
                                (n − r)!



n is the number of different items and r is the number of items taken
                               at a time
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!



                  4
                      P2
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!
                   P =
                  4 2
                       (4 − 2)!
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!     4!
                   P =
                  4 2          =
                       (4 − 2)! 2!
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!     4! (4)(3)(2)(1)
                   P =
                  4 2          =   =
                       (4 − 2)! 2!     (2)(1)
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!     4! (4)(3)(2)(1)
                   P =
                  4 2          =   =
                       (4 − 2)! 2!     (2)(1)
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!     4! (4)(3)(2)(1)
                   P =
                  4 2          =   =
                       (4 − 2)! 2!     (2)(1)
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!     4! (4)(3)(2)(1)
                   P =
                  4 2          =   =             =12
                       (4 − 2)! 2!     (2)(1)
Example 2
Four letters are taken two at a time. This is the permutation of finding as
              many two-letter groupings of the four letters.

                                      n!
                                P=
                               n r
                                   (n − r)!


                          4!     4! (4)(3)(2)(1)
                   P =
                  4 2          =   =             =12
                       (4 − 2)! 2!     (2)(1)


                  12 two-letter groupings can be made
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!



                   6
                       P3
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!


                          6!
                    P =
                   6 3
                        (6−3)!
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!


                          6!    6!
                    P =
                   6 3        =
                        (6−3)! 3!
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!


                          6!    6!
                    P =
                   6 3        =    = (6)(5)(4)
                        (6−3)! 3!
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!


                          6!    6!
                    P =
                   6 3        =    = (6)(5)(4) =120
                        (6−3)! 3!
Example 3
How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and
                   9 if no repetitions are allowed?


                                     n!
                               P=
                              n r
                                  (n − r)!


                          6!    6!
                    P =
                   6 3        =    = (6)(5)(4) =120
                        (6−3)! 3!



                   120 3-digit number can be formed
Understanding Break
         1. What is the short way of writing (6)(5)(4)(3)(2)(1)?


                     2. Is (2!)(3!) the same as 6! ?


              3. Which of the following is the value of 0! ?
                     a. 0 b. 1 c. -1 d. ±1
       4. Which of the following are examples of permutations?
             a. Arranging 5 people on a bench that seats 5
b. Choosing 4 letters from 6 and writing all possible arrangements of the
                                choices
            c. Selecting a committee of 3 from 10 women
Understanding Break
         1. What is the short way of writing (6)(5)(4)(3)(2)(1)?
                                     6!
                      2. Is (2!)(3!) the same as 6! ?


              3. Which of the following is the value of 0! ?
                     a. 0 b. 1 c. -1 d. ±1
       4. Which of the following are examples of permutations?
             a. Arranging 5 people on a bench that seats 5
b. Choosing 4 letters from 6 and writing all possible arrangements of the
                                choices
            c. Selecting a committee of 3 from 10 women
Understanding Break
         1. What is the short way of writing (6)(5)(4)(3)(2)(1)?
                                     6!
                      2. Is (2!)(3!) the same as 6! ?
              No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1)
             3. Which of the following is the value of 0! ?
                      a. 0 b. 1 c. -1 d. ±1
       4. Which of the following are examples of permutations?
             a. Arranging 5 people on a bench that seats 5
b. Choosing 4 letters from 6 and writing all possible arrangements of the
                                choices
            c. Selecting a committee of 3 from 10 women
Understanding Break
         1. What is the short way of writing (6)(5)(4)(3)(2)(1)?
                                     6!
                      2. Is (2!)(3!) the same as 6! ?
              No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1)
             3. Which of the following is the value of 0! ?
                      a. 0 b. 1 c. -1 d. ±1
       4. Which of the following are examples of permutations?
             a. Arranging 5 people on a bench that seats 5
b. Choosing 4 letters from 6 and writing all possible arrangements of the
                                choices
            c. Selecting a committee of 3 from 10 women
Understanding Break
         1. What is the short way of writing (6)(5)(4)(3)(2)(1)?
                                     6!
                      2. Is (2!)(3!) the same as 6! ?
              No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1)
             3. Which of the following is the value of 0! ?
                      a. 0 b. 1 c. -1 d. ±1
       4. Which of the following are examples of permutations?
             a. Arranging 5 people on a bench that seats 5
b. Choosing 4 letters from 6 and writing all possible arrangements of the
                                choices
            c. Selecting a committee of 3 from 10 women
Understanding Break
         1. What is the short way of writing (6)(5)(4)(3)(2)(1)?
                                     6!
                      2. Is (2!)(3!) the same as 6! ?
              No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1)
             3. Which of the following is the value of 0! ?
                      a. 0 b. 1 c. -1 d. ±1
       4. Which of the following are examples of permutations?
             a. Arranging 5 people on a bench that seats 5
b. Choosing 4 letters from 6 and writing all possible arrangements of the
                                choices
            c. Selecting a committee of 3 from 10 women
Homework
Homework




                      p. 174 #1-25 odd




"The inner fire is the most important thing mankind possesses."
                        - Edith Sodergran

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Integrated Math 2 Section 4-6

  • 2. Essential Question How do you find the number of permutations of a set? Where you’ll see this: Cooking, travel, music, sports, games
  • 4. Vocabulary 1. Permutation: An arrangement of items in a particular order 2. Factorial: 3. n Pr :
  • 5. Vocabulary 1. Permutation: An arrangement of items in a particular order ORDER IS IMPORTANT!!! 2. Factorial: 3. n Pr :
  • 6. Vocabulary 1. Permutation: An arrangement of items in a particular order ORDER IS IMPORTANT!!! 2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)... (3)(2)(1) 3. n Pr :
  • 7. Vocabulary 1. Permutation: An arrangement of items in a particular order ORDER IS IMPORTANT!!! 2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)... (3)(2)(1) 5! = (5)(4)(3)(2)(1) 3. n Pr :
  • 8. Vocabulary 1. Permutation: An arrangement of items in a particular order ORDER IS IMPORTANT!!! 2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)... (3)(2)(1) 5! = (5)(4)(3)(2)(1) = 120 3. n Pr :
  • 9. Vocabulary 1. Permutation: An arrangement of items in a particular order ORDER IS IMPORTANT!!! 2. Factorial: The number of permutations of n items: n(n - 1)(n - 2)... (3)(2)(1) 5! = (5)(4)(3)(2)(1) = 120 3. n Pr : n permutations taken r at a time; n is the number of different items and r is the number of items taken at a time
  • 10. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed?
  • 11. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed?
  • 12. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed? 3
  • 13. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed? 3 2
  • 14. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed? 3 2 1
  • 15. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed? 3 2 1 (3)(2)(1)
  • 16. Example 1 How many 3-digit numbers can be formed from the digits 4, 7, and 9 if no repetitions are allowed? 3 2 1 (3)(2)(1) = 6 numbers
  • 18. Permutation Formula n! n Pr = (n − r)!
  • 19. Permutation Formula n! n Pr = (n − r)! n is the number of different items and r is the number of items taken at a time
  • 20. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters.
  • 21. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)!
  • 22. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4 P2
  • 23. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! P = 4 2 (4 − 2)!
  • 24. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! 4! P = 4 2 = (4 − 2)! 2!
  • 25. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! 4! (4)(3)(2)(1) P = 4 2 = = (4 − 2)! 2! (2)(1)
  • 26. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! 4! (4)(3)(2)(1) P = 4 2 = = (4 − 2)! 2! (2)(1)
  • 27. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! 4! (4)(3)(2)(1) P = 4 2 = = (4 − 2)! 2! (2)(1)
  • 28. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! 4! (4)(3)(2)(1) P = 4 2 = = =12 (4 − 2)! 2! (2)(1)
  • 29. Example 2 Four letters are taken two at a time. This is the permutation of finding as many two-letter groupings of the four letters. n! P= n r (n − r)! 4! 4! (4)(3)(2)(1) P = 4 2 = = =12 (4 − 2)! 2! (2)(1) 12 two-letter groupings can be made
  • 30. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed?
  • 31. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)!
  • 32. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)! 6 P3
  • 33. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)! 6! P = 6 3 (6−3)!
  • 34. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)! 6! 6! P = 6 3 = (6−3)! 3!
  • 35. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)! 6! 6! P = 6 3 = = (6)(5)(4) (6−3)! 3!
  • 36. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)! 6! 6! P = 6 3 = = (6)(5)(4) =120 (6−3)! 3!
  • 37. Example 3 How many 3-digit numbers can be formed from the digits 4, 5, 6, 7, 8, and 9 if no repetitions are allowed? n! P= n r (n − r)! 6! 6! P = 6 3 = = (6)(5)(4) =120 (6−3)! 3! 120 3-digit number can be formed
  • 38. Understanding Break 1. What is the short way of writing (6)(5)(4)(3)(2)(1)? 2. Is (2!)(3!) the same as 6! ? 3. Which of the following is the value of 0! ? a. 0 b. 1 c. -1 d. ±1 4. Which of the following are examples of permutations? a. Arranging 5 people on a bench that seats 5 b. Choosing 4 letters from 6 and writing all possible arrangements of the choices c. Selecting a committee of 3 from 10 women
  • 39. Understanding Break 1. What is the short way of writing (6)(5)(4)(3)(2)(1)? 6! 2. Is (2!)(3!) the same as 6! ? 3. Which of the following is the value of 0! ? a. 0 b. 1 c. -1 d. ±1 4. Which of the following are examples of permutations? a. Arranging 5 people on a bench that seats 5 b. Choosing 4 letters from 6 and writing all possible arrangements of the choices c. Selecting a committee of 3 from 10 women
  • 40. Understanding Break 1. What is the short way of writing (6)(5)(4)(3)(2)(1)? 6! 2. Is (2!)(3!) the same as 6! ? No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1) 3. Which of the following is the value of 0! ? a. 0 b. 1 c. -1 d. ±1 4. Which of the following are examples of permutations? a. Arranging 5 people on a bench that seats 5 b. Choosing 4 letters from 6 and writing all possible arrangements of the choices c. Selecting a committee of 3 from 10 women
  • 41. Understanding Break 1. What is the short way of writing (6)(5)(4)(3)(2)(1)? 6! 2. Is (2!)(3!) the same as 6! ? No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1) 3. Which of the following is the value of 0! ? a. 0 b. 1 c. -1 d. ±1 4. Which of the following are examples of permutations? a. Arranging 5 people on a bench that seats 5 b. Choosing 4 letters from 6 and writing all possible arrangements of the choices c. Selecting a committee of 3 from 10 women
  • 42. Understanding Break 1. What is the short way of writing (6)(5)(4)(3)(2)(1)? 6! 2. Is (2!)(3!) the same as 6! ? No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1) 3. Which of the following is the value of 0! ? a. 0 b. 1 c. -1 d. ±1 4. Which of the following are examples of permutations? a. Arranging 5 people on a bench that seats 5 b. Choosing 4 letters from 6 and writing all possible arrangements of the choices c. Selecting a committee of 3 from 10 women
  • 43. Understanding Break 1. What is the short way of writing (6)(5)(4)(3)(2)(1)? 6! 2. Is (2!)(3!) the same as 6! ? No; (2)(1)(3)(2)(1) ≠ (6)(5)(4)(3)(2)(1) 3. Which of the following is the value of 0! ? a. 0 b. 1 c. -1 d. ±1 4. Which of the following are examples of permutations? a. Arranging 5 people on a bench that seats 5 b. Choosing 4 letters from 6 and writing all possible arrangements of the choices c. Selecting a committee of 3 from 10 women
  • 45. Homework p. 174 #1-25 odd "The inner fire is the most important thing mankind possesses." - Edith Sodergran

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