Permutations

4! is read as 4 factorial

and is...

4 x 3 x 2 x 1.  It equals

24.

3! =

5! is read as 5 factorial

and is...

5 x 4 x 3 x 2 x 1.  It

equals 120.

7! = 7 x 6 x 5 x     

3! =         x        x 

0! is read as 0 factorial

and equals 1.

3! is read as 3 factorial

and is...

3 x 2 x 1.  It equals 6.

4! =       x      x     x

Multiplication Principle:
If you want to chose one element from a set A containing n elements and one element from a set B containing m elements and one element from a set C containing p elements and so on, Then the number of ways in which you can choose these elements are given by the formula: 
 =n × m × p × ...
Example1
You and a group of friends love to go to the movies.You always purchase popcorn, soda, and candy. You can choose from buttered or unbuttered popcorn, five different kinds of soda, and 18 different kinds of candy.  Determine the number of possible configurations.
x
x
=
Example2
You have three pairs of jeans, five t-shirts, and two pairs of shoes.  Determine the number of possible ways to dress up.
x
x
=

I have two different rings, four

hats, and three jackets. How

many ways can I select a ring,

hat, and jacket to wear?

9

 24

1/24

 

12

 

Susan has 4 colors of wrapping paper,

3 colors of ribbon, and 2 types of tape.

How many different ways does she have

to wrap a package with one color of

wrapping paper, one color of ribbon,

and one type of tape?

1
4
12
24
Create a tree diagram for the following problem. An ice cream shop has vanilla and
chocolate ice cream, and chocolate and strawberry syrup.  You can also choose
between sprinkles or nuts. Make a tree diagram showing the outcomes.
Ice Cream Flavor     Syrup Flavor                Topping
vanilla











chocolate





chocolate





strawberry
nuts


sprinkles


sprinkles

nuts
Permutations
Let E be a set containing n elements ( Card(E) =n ), Let  rbe a natural number  ( r ≤ n). Then a permutation without repetition of the r elements selected from E is any ordered sequence taken from the elements of E containing exactly r distinct elements(No repeated element in a single permutation)
Eg: Let E= {1,2 ,3 ,4,5,6} then (1,3,5) is a permutation of E containing 3 elements, (1,5,3) is another permutation of E containing 3 elements, (1,2) is a permutation of E containing two element and (652413) is a permutation of E containing 6 elements>Note1: (1,1,3) and (2,2) are not permutations without repetitions> Note2: Permutations without repetitions are simply called permutations

Permutations - order matters no repetition 

 

Of n things taken r at a time.

n

P

r

=           


or

n count down r times

(n-r)!

n!

N is the number of items being arranged in groups

 of r elements each.

  

How many items are there to pick from ?

n

P

r

=

           

or

n count down r times

(n-r)!

n!

N is the number of items being arranged in lots of r.

 

 

How many items are in each group?         items

n

P

r

=  

         

or

n count down r times

(n-r)!

n!

For this problem start at 6 and count down two place.

Permutations - order matters no repetition   

Of n things taken r at a time.

n

P

r

6 x 5

n multiplied by count down r places

(n-r)!

n!

or

6

P

2

(6-2)!
6!

Permutations - order matters no repetition  

Of n things taken r at a time.

n

P

r

6 x 5 x 4 =

=   

               or

n count down r times

(n-r)!

n!

6

P

3

(6-3)!
6!

Permutations - order matters no repetition  

Of n things taken r at a time.

n

P

r

7 x 6 x 5 =

=  

               or

n count down r times

(n-r)!

n!

7

P

3

(7-3)!
7!

Permutations - order matters no repetition  

Of n things taken r at a time.

8

P

3

   x     x      =  336

count down 3 spots
(8-3)!
8!

8!
!
Ex:1 How many anagrams are there of the letter SEA?
(An anagram is a word or phrase formed by rearranging the letters of a different word, typically using all the original letters exactly once)
Applications
The possible cases are :  
SEA, SAE, ESA, EAS, ASE , AES
Ex:2
Four horses are running a race.  How manydifferent ways can these horses come in first, second, and third?
You forget the pin number for your computer which consists of 4 distinct digits. You remember that the firstdigit is 1 and the remaining three digits are all even number.What is the maximum number of attempts before you can find the correct pin and unlock your computer?
Rule
Answer
r-lists
Repetition is allowed and the order is important 
Formula: If a set contains n elements and you want to choose r elements each time then the number of ways in which this can be done is: 
n ×n×n×...×n = nr
Direction: Drag each text to its correct place
Repetition not allowed - order important
Repetition Allowed - order important 
r-lists
?
Permutation
?
Permutation
r-lists
nr
?
n!/ (n-r)!
?
Example1How many 2-digit numbers containing only the digits 3,4,5,6 are there?
Examples of these numbers are 33, 45, 54, 55 ... Here we have the set {3,4,5,6} and we want to chose 2elements of it with repetition and the order is of courseimportant (since 34 ≠43 ).So we must use the r-lists:
Here we have: 
Then, there are
n=
nr=
and 
=
r = 
The answer can be found using multiplication principal:
The numbers are as follows: 
33, 34, 35, 3643, 44, 45, 4653, 54, 55, 5663, 64, 65, 6616 numbers 
We must chose two digits to fill in the two boxes:
The first box can be filled with 4 digits and the secondcan be filled with 4 digits also, so the is 4×4 = 16
Example2:Suppose we have a coin (one face is called head (H) and the other is called tail (T) ). We tossedthis coin three times in air and we noted its upper face in each time: such as THT which means we obtained Tail then Head then Tail respectively. Question: How many different outcomes we have?
Solution
we have to choose 3 elements from the letters {H, T}The order is important and we must have repetitions. So: 
Then nr=
n=
and
r=
We can think of this question in the following way:
We have to fill the three boxes letters {H or T}:
So we have 2 choices in the first box, 2 choices in the 
second box and 2 choices in the third box, 
Using multiplication principal: 
The number of outcomes is : 2×2×2 = 23=8
H
T
T
H
H
T
To find all the previous outcomes, we use a Tree-diagram:

Ex:1
In a certain state, each automobile license platenumber consists of two english letters followed by afour-digit number.  How many distinct license plate numbers can be formed?
In a certain Country, each automobile license plate consists of three digits followed by three English Capital letters.  How many distinct license plate numbers can be made with no repeat letters.
Ex:2
×
=
Ex: 3
How many four-digit numbers can be formed?
Ex: 4
How many four digit numbers can be formed 
with distinct digits?
Ex: 5
How many four digit numbers can be formed 
less than 5000?
Ex: 6
How many ways can six people sit in a six-passenger car? 
A minivan has 8 seats (including the driver seat).
A group of 8  friends want to go to a trip in this
 minivan. How many possible ways they can sit in 
this minivan if there are only four  drivers among them?
Ex: 7
×
=
A symbol can be an English letter (Upper or lower case), a digit from 0 to 9 or one of : %, $, @, -, _, # You Know that the password for your account is formed of 6 symbols.How many possible passwords are there? ( Note: the letters are case: sensitive: means that A≠a )
Ex:8
Rule
Rule
Ex:9
Answer
Rule
Answer
Note:If you want something AND something use     (×)
Note:If you want something OR something use     (+)
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