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```					Fundamental counting principle
Factorials
Permutations
Combinations
Fundamental counting principle
Fundamental Counting Principal = Fancy
way of describing how one would
determine the number of ways a sequence
of events can take place.
Fundamental counting principle
You are at your school cafeteria that allows you to choose a
lunch meal from a set menu. You have two choices for the
Main course (a hamburger or a pizza), Two choices of a drink
(orange juice, apple juice) and Three choices of dessert (pie,
ice cream, jello).
12 meals
How many different meal combos can you select?_________
Method one: Tree diagram       Lunch
Hamburger                            Pizza

Apple          Orange             Apple          Orange

Pie              Pie             Pie            Pie
Icecream         Icecream        Icecream       Icecream
Jello            Jello           Jello          Jello
Fundamental counting principle
Method two:   Multiply number of choices

2 x 2 x 3 = 12 meals

Ex 2: No repetition
During the Olympic 400m sprint, there are 6 runners. How
many possible ways are there to award first, second, and
third places?
1st    2nd      3rd
3 places       6       5        4
____ x ____ x ____ = 120 different ways
Fundamental counting principle
Ex 3: With repetition
License Plates for cars are labeled with 3 letters followed by 3
digits. (In this case, digits refer to digits 0 - 9. If a question
asks for numbers, its 1 - 9 because 0 isn't really a number)
How many possible plates are there? You can use the same
number more than once.

26 26 26 10 10 10
___ x ___ x ___ x ___ x ___ x ___ = 17,576,000 plates

Ex 4: Account numbers for Century Oil Company consist
of five digits. If the first digit cannot be a 0 or 1, how many
account numbers are possible?

___ x ___ x ___ x ___ x ___ = 80,000 different account #’s
8     10    10 10       10
Factorials
5 • 4 • 3 • 2 • 1 = 5! Factorial
7!= 7 • 6 • 5 • 4 • 3 • 2 •1 = 5040

42

56
Permutations
Permutations = A listing in which order IS important.

Can be written as:    P(6,4)        or     6P4

P(6,4) Represents the number of ways 6 items can be
taken 4 at a time…..

Or 6 x 5 x 4 x 3 = 360          Or 6 (6-1) (6-2) (6-3)

2730
Find P(15,3) = _____

15 x 14 x 13
Permutations - Activity

Write the letters G R A P H on the top of your paper.

Compose a numbered list of different 5 letter Permutations.
-(not necessarily words)

On the bottom of your paper write how many different
permutations you have come up with.

Hint: You may wish to devise a strategy or pattern for
finding all of the permutations before you start.
Permutations
Use the same formula from section 52 to solve these WPs.

Ex1. Ten people are entered in a race. If there are no ties,
in how many ways can the first three places come out?
10     9     8
___ x ___ x ___ = 720

Ex2. How many different arrangements can be made with
the letters in the word LUNCH?
5! or           4     3     2    1
___ x ___ x ___ x ___ x ___ = 120
5

Ex3. You and 8 friends go to a concert. How many
different ways can you sit in the assigned seats?

9! = 362,880
Combinations
Combinations = A listing in which order is NOT important.

Can be written as:    C(3,2)          or    3C2

C(3,2) means the number of ways 3 items can be
taken 2 at a time. (order does not matter)

Ex. C(3,2) using the letters C A T

CA CT AT

n = total
r = What you want
Combinations
n = total
r = What you want

7x6           42    = 21
C(7,2)                                =
2x1           2

Which is not an expression for the number of ways 3 items
can be selected from 5 items when order is not considered?
Combinations
Permutations = Order IS important

8     7     6
P(8,3) = ___ x ___ x ___ = 336

Combinations = Order does not matter

C(8,3) =                              56
Combinations
Ex1. A college has seven instructors qualified to teach a
special computer lab course which requires two instructors
to be present. How many different pairs of teachers could
there be?

C(7,2) =           21

Ex2. A panel of judges is to consist of six women and three
men. A list of potential judges includes seven women and six
men. How many different panels could be created from this
list?
Women                        Men
C(7,6)                       C(6,3) =            20

7
7*20 = 140  140 choices

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