Finite Math Counting Arrangements: Permutations http://finitehelp.com/

1. 2.2
Counting Arrangements: Permutations

2. Counting Arrangements
 The idea of counting arrangements is to find the
number of different ways to arrange a set of
things.
 For example, take the set {1, 2, 3}. There are 6
ways to arrange this set into a 3-digit number
without using a number twice:
123
132
213
231
312
321
 Each of these arrangement is called a
“permutation.”

3. Permutation
 In a permutation, order of the arrangement
matters. That means 123 is a different
arrangement than 321. More on that in a later
section.
 Also, in a permutation you cannot use the same
element twice.

4. Permutation Formula

n – number of things you’re choosing from
r – number you’re choosing
(if you are unfamiliar with the “!” notation, check
out the factorial tutorial)
 With our previous example, n = 3 (set has 3
elements, 1, 2, 3), and r = 3 (3-digit number).
 P(3, 3) = 3!/(3-3)! = 3!/0! = 3 x 2 x 1 = 6

5. Slot Method
 To calculate the number of permutations, the
formula is given to you. However, if you
recognize a problem to be a permutation
problem, the “slot method” is a simpler way.
 Let’s take our previous example – 3-digit
number from {1, 2, 3}. Since it’s a 3-digit
number, there are 3 slots:

6. The number on digit 1 represents
the number of elements “permitted”
on that slot. In this case, there are
3.
(1, 2, 3 can be the first digit)
Since there can be no repeats, there
are only two elements permitted on
digit 2:
And only one left for digit 3.
Simply multiply for the final answer: 3 x 2 x 1 =
6.

7. Slot Method for
Multiplication Principle
 Slot method can be applied for multiplication
principle problems as well [1.4]. Lets take the
Happy Meal example (3 entrées, 3 drinks, 4
toys):
 3 x 3 x 4 = 36.

8. Quiz 2.2 #1
 How many 3-letter permutations can be formed
with the set {A, B, C, D, E}?
[Try using both the formula and slots]
A. 20
B. 60
C. 120

9. Quiz 2.2 #1
 How many 3-letter permutations can be formed
with the set {A, B, C, D, E}?
[Try using both the formula and slots]
A. 20
B. 60
C. 120
 Answer: B

10. The letter permutation problem
 Lets consider the word “infinite.” How many 8-
letter arrangements can be formed from the
letters of this word?
 This is a special problem that requires its own
trick. Here’s how it goes:
 First, treat it as a normal 8-slot permutation:
8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

11.  However, that’s too many, because some letters
are repeated.
 So, we divide it by the factorial of number of
times a letter is repeated.
 For example, “i” is repeated 3 times, so you
divide by 3!. “n” is repeated twice, so divide by
2!:
3 “i”s 2 “n”s
 Therefore, the answer is 40320/(6*2) = 3360

12. Quiz 2.2 #2

13. Quiz 2.2 #2
 How many 6 letter arrangements can be formed
from the word “effect”?
A. 30
B. 60
C. 180

14. Quiz 2.2 #2
 How many 6 letter arrangements can be formed
from the word “effect”?
A. 30
B. 60
C. 180

15. Quiz 2.2 #2
 How many 6 letter arrangements can be formed
from the word “effect”?
A. 30
B. 60
C. 180
 Answer: C

16. Summary
 Definition:
 Permutation (order matters, no repetition)
 How to find the number of permutations
 Formula:
 Slot method
 The letter permutation problem

17.  Features
 27 Recorded Lectures
 Over 116 practice problems with recorded solutions
 Discussion boards/homework help
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