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
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
