Download m150cn-jm2

Chapter 2
Section 2.3
Introduction to Combinatorics
Counting
It is often important to be able to count how many elements are in a set. This
seems strange when it is first discussed since you usually think just enumerate
(count them one by one) the elements in the set.
For example: The number of elements in set A = {, , , }
1 2 3 4
n(A)=4
The problem with this method is that sometimes sets are too big (in the millions
or billions) or very complicated and are described rather than having the
elements listed out. One way to approach this is with a decision tree.
Example: A certain style of T-shirt comes in 3 different colors, red (R), green (G),
blue (B) and 4 different sizes, small (S), medium (M), large (L) and extra large
(XL). How many different types of this style of T-shirt are there?
Answer: 12
R
G
B
S
M
L
XL
S
M
L
XL
S
M
L
XL
S
M
L
XL
S
M
L
XL
S
M
L
XL
The idea is to think of the
tree to the right in levels.
Each level represents a
decision that needs to be
made or a bunch of choices
or options you have.
G
B
Choose a Color (3 choices)
S M L XL
S M L XL
Choose a Size (4 choices)
R
S M L XL
What we notice is if multiply the
number of choices or decisions
together we get the total number of
different types of T-shirt.
3
4
Color
Size
34 = 12
Total Types
This idea can be generalized as described below.
Fundamental Counting Principle
If an element in a set can be described with a series of k choices (decisions or
options) with the number of different options for each of the k choices being:
n1,n2,n3,…,nk respectively. The total number of elements in the set is:
n1n2n3  nk
Fundamental Counting Principle
The idea for counting the number of things in a set by breaking it into a series of
choices, decisions or options to make then multiplying them together is called
the Fundamental Counting Principle.
This says if there are m choices for how to select option 1, along with n choices for
how to select option 2 then the number of ways to select both option 1 and option 2
at the same time is mn. (If there are more than two options you would multiply
more numbers together.)
The tough part of any idea like this is how to apply it. Consider the following
example.
A fast food restaurant offers a combo special. You get a choice from each category
below. How many ways can you order a combo special.
Sandwich
Hamburger
Cheeseburger
Chicken
Fish
4 choices
Side
French Fries
Onion Rings
Side Salad
3 choices
Drink
Cola
Diet Cola
Root Beer
Orange
Lemonade
5 choices
The Fundamental Counting
Principle shows the total
ways to order a combo
special is:
4  3  5 = 60
The difficult thing is how to apply this idea.
A certain issue of a series of car license plates in the state of Ohio had the plates
consisting of 3 letters followed by 4 numbers. For example: CGT 4317. How many
different license plates can be made this way?
We use the Fundamental Counting Principle. There are 7 choices to be made, the
1st letter, 2nd letter, 3rd letter, 1st number, 2nd number, 3rd number and 4th number.
26
26
26
1st
2nd
3rd
letter
letter
letter
10
10
10
10
1st
2nd
3rd
4th
number
number
number
number
26262610101010
= 263104
= 175,760,000
Some restrictions on the elements may limit some of your choices.
In the example above how many license plates begin with the letter P?
1262610101010
1
26
26
10
10
10
10
= 262104
1st
2nd
3rd
1st
2nd
3rd
4th
= 6,760,000
letter
letter
letter
number
number
number
number
How many begin with a vowel?
5
26
26
1st
2nd
3rd
letter
letter
letter
10
10
10
10
1st
2nd
3rd
4th
number
number
number
number
5262610101010
= 5262104
= 33,800,000
The choices you make at one stage may restrict the number of choices you make at
the next stage.
For example:
If 10 horses run a race in how many different ways can they finish in 1st, 2nd, and 3rd
place? (In racing this is called a trifecta.)
Whichever horse finished
10
9
8
st can not finish 2nd and
1098 = 720
1
1st
2nd
3rd
neither of them can finish
place
place
place
3rd.
In how many different ways can all 10 horses finish the race in places 1 through 10?
Answer: 10987654321 = 3,628,800 = 10! (read "ten factorial")
The factorial is a shorthand way of writing a long string of consecutive
decreasing numbers being multiplied together. Here are some more examples.
0! = 1
4! = 4321 = 24
8! = 87654321 = 40,320
1! = 1
5! = 54321= 120
9! = 987654321 = 362,880
2! = 21 = 2
6! = 654321 = 720
10! = 10987654321 = 3,628,800
3! = 321=6
7! = 7654321 = 5040
11! = 1110987654321=39,916,800
Evaluating Expressions with Factorials
When finding the value of an expression with a factorial (!) in it remember a couple of
tips.
1. In fractions usually much cancellation can occur
2. Remember to do what is in parenthesis first.
Find the value of each expression below.
a)
10!
7!
10! 10  9  8  7  6   2 1

 10  9  8  720
7!
7  6    2 1
b)
50!
(50  2)!
50!
50! 50  49  48  47  2 1


 50  49  2450
(50  2)! 48!
48  47  2 1
12!
c)
3!(12  3)!
12!
12! 12 1110  9  8  2 1 12 1110



 2 1110  220
3!(12  3)! 3!9!
3  2 1  9  8    2 1
3  2 1