1. Accelerated Precalculus
How do I love permutations ? . . .
Let me count the ways.
(With apologies to Elizabeth Barrett Browning)

2. One-Minute Question
If m AE = 100 and mC = 20, find m BD. A B C E D

3. Ready to start?
Watch carefully.
Take good notes.

4. And Don’t Panic...
Ask Questions if things aren’t clear!

5. 4  3  2  1 = 24 The letters M , A , T , and H are to be used
Problem:
The letters M , A , T , and H are to be used
to create a password to log onto the computer.
If no letter is to be repeated, how many different four-letter passwords could be created?
Solution: First, draw blanks.
4  3  2  1 = 24
_____ _____ _____ _____
Then, count how many choices there are for each selection.

6. Definition (to know and use):
Factorial:
If n is a positive integer, then n factorial, written as n!, is computed as:
n! = n  (n-1)  (n-2)   3  2  1

7. Now try these:
720
362,880 6 1
120
6! =
9! =
3! =
1! =
5! =

8. Special Case Definition:
I realize zero is not a positive integer, and,
therefore, does not fit the definition for a factorial,
but we will need 0! to be defined as such:
0! = 1

9. Problem:
From a stack of 9 different books, 4 books will be placed on a shelf. How many different arrangements are possible?
Solution: Again, draw blanks.
9  8  7  6 = 3024
_____ _____ _____ _____
Then, count how many choices there are for each selection.

10. Definition (to know and use):
Permutation:
An arrangement of things in which order is important. The number of permutations of n objects chosen r at a time, is computed as:
P(n,r) = nPr = ___n!___
(n-r)!

11. Solution: Use the formula this time.
This was a permutation!!
From a stack of 9 different books, 4 books will be placed on a shelf. How many different arrangements are possible?
Solution: Use the formula this time.

12. Now try these: 20 10
360
120 1
P(5, 2) =
P(10, 1) =
P(6, 4) =
5P5 =
5P0 =

13. 7 6 5 4 3 2 1 = 420 (3  2  1)  (2  1) But Wait!!
Problem:
How many different arrangements can be made using all the Scrabble tiles:
A A A P P R K ?
But Wait!!
If we switch the places of the A’s - will anyone notice??
Solution: Again, draw blanks.
7 6 5 4 3 2 1 =
420
__ __ __ __ __ __ __
(3  2  1)  (2  1)
Then, count how many choices there are for each selection.
*But then, divide by the number of ways each repeated element can be arranged. (3! for the As and 2! for the Ps)

14. Problem:
A group of 6 children are forming a circle to play “Ring Around the Rosie” ? In how many ways can they join hands?
But Wait!
As soon as they start moving, won’t some of these ways be repeated?
Solution: Again, draw blanks.
6  5  4  3  2  1=
__ ___ ___ __ ___ __
120 6
Then, count how many choices there are for each selection.
*But then divide by the number of places to which the first person could move.

15. Classwork!

16. Assignment:
Pg. 634: 3, 9 – 25 odd, 35 – 39 odd, 43, 46, 55 – 81 odd,
Work together on this and ask each other questions!