**Finite Mathematics Lesson 6**

**Chapters 5.4 - 5.5**

This may be a little harder. I recommended writing out sample spaces by hand or even drawing pictures to help you see how to count.

* Counting*Counting is a general term that describes a
method for counting objects in a set. We may want to count the number of objects in
a set or the numbers of ways the objects can be arranged. We could also count the
number of possible poker hands that could be dealt given a single deck of cards.
Many of these type of counting problems are much to difficult to use a trail and error
approach. Could you image actually trying to count out each possible hand of
poker?. Chapter 5 will explore different counting techniques.

**Course Notes**

**Section 5.4 The Multiplication
Principle**

*Generalized Multiplication Principle
*Suppose a task
consists of

**Example**

A women has 4 blouses, 3 skirts and and 5 pairs of shoes. Assuming
the woman does not care what she looks like, how many different outfits can she
wear?

- She has 4 × 3 × 5 = 60 different outfits.

**Example - A classic
Problem**

License plates for cars have to be unique. If a license plate contains 6
characters: 2 letters followed by 4 digits i.e. we2354 etc..

- There are 26 × 26 × 10 × 10 × 10 × 10 = 6,760,000 different plates.

What if letters were allowed to repeated but numbers were not allowed to be repeated?

- 26 × 26 × 10 × 9 × 8 × 7 = 3,407,040 different plates

**Example
**Fred has 10 different pairs of shoes. In
how many ways can Fred put on a pair of shoes that

**Let's look at this problem
two ways....The first way is to count all the combinations. **

- Let's name each pair A, B, C, ...., J
- Every pair has a left
and a
**right**shoe. - Assume we have to wear a
**left**and a**right**. - A
**A**, B**B**, C**C**, D**D**, E**E**, F**F**, G**G**, H**H**, I**I**, J**J** - Start the left shoe A,
A can be paired with
**right**shoes starting with**B.**That means shoe can be paired with**9**other**left**shoes. - Next take the
**left**shoe**B**, it has already been paired with A so that means B can be paired with**8**remaining**right**shoes.**C**,**D**,...,**J** - Continue this pattern until you reach the
last shoe.
**Left**

Shoe**mismatching**

pairs**Right**

Shoe**mismatching**

pairsA 9 other shoes **A**9 other shoes B 8 **B**8 C 7 **C**7 D 6 **D**6 E 5 **E**5 F 4 **F**4 G 3 **G**3 H 2 **H**2 I 1 **I**1 J Already paired **J**Already paired Total **45****45****There are 90 mismatching pairs.**

**The second way is much
easier, but harder conceptionally. Apply the multiplication rule...**

**Each pair**can mate with 9 different pairs.- Start the Pair A
**A**, it can be paired with 9 other pairs excluding itself. - Next choose Pair B
**B**, it can be paired with 9 other pairs. In this case, we don't get a repeat because there are 2 combinations for each pairing. A**B**and B**A**are different mismatching pairs.

- There are 10 × 9 = 90 different mismatching pairs.

When you are ready, click here for the assignment for this section.

**Section 5.5 Permutations and
Combinations**

*Permutation**P*(*n,r*) = the number of permutations of*n*objects taken*r*at a time.

**Example**Order is important when talking about permutations. Lets say 3 people (**Andy**,**Betty**, and**Clark**) compete in a running race. How many possible outcomes are there? 6 distinct outcomes**1st Place****2nd Place****3rd Place****Andy****Betty****Clark****Andy****Clark****Betty****Betty****Clark****Andy****Betty****Andy****Clark****Clark****Andy****Betty****Clark****Betty****Andy**Here, order is very important (especially to the

If awards (1st place and 2nd place) were given to the top 2 finishers, how many different arrangements of awards could be given?

*P*(**3***,*2)

**Here is the formula.**The first number is the total number of objects,*P*(*n,r*) =*n*(*n*- 1)(*n*- 2) · · · (*n*-*r*+ 1)

.**n**

The last number is**1**more than*n*-*r*If you were to choose 3 objects from 10 possible choices (order being important)

The first numbers 10 and the last number is 10 - 3 +1 = 7 +1 = 8

*P(10,3) =*10 · 9 · 8 =720

C(*Combination**n,r*) = the number of combinations of*n*objects taken*r*at a time.

**Example**Order is**NOT**important when talking about combinations. Lets say 3 people (**Andy**,**Betty**, and**Clark**) compete in a running race. Here are the possible outcomes.**1st Place****2nd Place****3rd Place****Andy****Betty****Clark****Andy****Clark****Betty****Betty****Clark****Andy****Betty****Andy****Clark****Clark****Andy****Betty****Clark****Betty****Andy**This time the top 2 finishers go on to the final round, how many different arrangements of finalists could be given?

**Here is the formula.****Notice there are less possibilities when order is not considered.**If you were to choose 3 objects from 10 possible choices (order

**NOT**important)

**These alternate forms for counting combinations and permutations are introduced in 5.7**You can use them now, I think the are easier to use.

An alternate form for *combinations*unordered **Notice***C*(7,4) =*C*(7,3) = 35 Why?An alternate form for *permutations*ordered **Notice***P*(7,4) = 840

**but***P*(7,3) = 210

A final note: Logically speaking, these are some of the hardest problems in the book. Don't be afraid to ask questions.

When you are ready, click here for the assignment for this section.

- There are additional online tutorials at the books' web site. http://cw.prenhall.com/bookbind/pubbooks/goldstein3/

Please notify me of any errors on this page joe@joemath.com