M11E321 Determine the number of permutations andor combinations or apply the fundamental counting principle Objectives Permutations Combinations Vocabulary A permutation is an arrangement of items in a particular order ID: 637140
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Slide1
6-7 Permutations & Combinations
M11.E.3.2.1: Determine the number of permutations and/or combinations or apply the fundamental counting principleSlide2
Objectives
Permutations
CombinationsSlide3
Vocabulary
A
permutation
is an arrangement of items in a particular order.
n factorialFor any positive integer n, n
! =
n
(
n – 1
) · … · 3 · 2 · 1
For
n
= 0,
n
! = 1Slide4
Vocabulary
Fundamental
Counting Principle
If
there are a ways the first event can occur and
b
ways the
second
event can occur, then there are
a x b
ways that both events can occur
.
Example:
You
are ordering dinner at a restaurant. You have a choice of soup
or salad
for an appetizer. A choice of steak, chicken, or tofu for a main entrée, and a choice of pie or ice cream for dessert. How many different meals can you have?Slide5
In how many ways can 6 people line up from left to right for a group photo?
Since everybody will be in the picture, you are using all the items from the
original set. You can use the
Multiplication Counting Principle
or
factorial notation
.
There are six ways to select the first person in line, five ways to select the
next person, and so on.
The total number of permutations is 6 • 5 • 4 • 3 • 2 • 1 =
6!
.
6!
= 720
The 6 people can line up in 720 different orders.
Finding PermutationsSlide6
Vocabulary
Permutations
If order
does
matter, then you are working with permutations.
The number of permutations of
n
items of a set arranged
r
items at the time is
n
P
r
Ex. Seven people are running a race. How many different outcomes of first, second, and third place are possible?
n = 7 (total number of items) and r = 3 (how many we’re choosing)
This is written 7P3Slide7
Permutations
n
Pr = for 0 ≤ r ≤ n
10
P
5
= =
= 5040Slide8
How many 4-letter codes can be made if no letter can be used twice?
Method 1:
Use the
Multiplication Counting Principle
.
26 • 25 • 24 • 23 = 358,800
Method 2:
Use the permutation formula. Since there are 26 letters
arranged 4 at a time,
n
=
26
and
r
=
4.
There are 358,800 possible arrangements of 4-letter codes with no duplicates.
26
P
4
= = = 358,800
26!
(
26
–
4
)!
26!
22!
Real World ExampleSlide9
Vocabulary
Combinations
If order
does
not
matter
, then you are working with permutations.
The number of
combination of
n
items of a set arranged
r
items at the time
isnCr
Ex. Seven people are running a race. How many different ways can 3 people receive a medal? (It doesn’t matter whether they get a gold or bronze)
n = 7 (total number of items) and r = 3 (how many we’re choosing)This is written
7C3Slide10
Combinations
n
Cr = for 0 ≤ r ≤ n
5
C
3
= =
= = 10Slide11
Evaluate
10
C4.
10 • 9 • 8 • 7
4 • 3 • 2 • 1
=
= 210
10
C
4
=
10!
4!(10 – 4)!
=
10!
4! • 6!
10 • 9 • 8 • 7 •
4 • 3 • 2 • 1 •
=
6
5
4
3
2
1
•
•
•
•
•
6
5
4
3
2
1
•
•
•
•
•
Finding CombinationsSlide12
A disk jockey wants to select 5 songs from a new CD that contains 12 songs. How many 5-song selections are possible?
Relate:
12 songs chosen 5 songs at a time
Define:
Let
n
= total number of songs.
Let
r
= number of songs chosen at a time.
You can choose five songs in 792 ways.
Use the
n
C
r
feature of your calculator.
Write:
n
C
r
=
12
C
5
Real World ExampleSlide13
A pizza menu allows you to select 4 toppings at no extra charge from a list of 9 possible toppings. In how many ways can you select 4 or fewer toppings?
The total number of ways to pick the toppings is
126 + 84 + 36 + 9 + 1 = 256.
There are 256 ways to order your pizza.
You may choose
4 toppings
,
3 toppings
,
2 toppings
,
1 toppings
, or
none
.
9
C
4
9
C
3
9
C
2
9
C
1
9
C
0
Real
Worl
Example