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CHAPTER 4
4-4:Counting
Rules
Instructor: Alaa saud
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•Counting Rules
• 1-Fundamental Counting Rule
• 2- Permutation
• 3- Combination
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1-Fundamental Counting Rule
 In a sequence of n events in which the first one
has k1 possibilities and the second event has k2 and
the third has k3, and so forth, the total number of
possibilities of the sequence will be
k1 · k2 · k3 · · · kn
Event 1
k1
Event 2
k2
……………
Event n
……………
kn
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Example 4-39:
A paint manufacturer wishes to manufacture several
different paints. The categories include
Color: red, blue, white, black, green, brown, yellow
Type: latex, oil
Texture: flat, semi gloss, high gloss
Use:
outdoor, indoor
How many different kinds of paint can be made if you can
select one color, one type, one texture, and one use?
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Solution :
Color
7
Type
.
2
Texture
.
3 .
Use
2
=84
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Q: A store manager wishes to display 8 different
brands of shampoo in a row. How many
different ways can this be done?
Q:How many ways 5-digit passwords can be
done ?
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 Factorial is the product of all the
positive numbers from 1 to a number.
n !  n  n  1 n  2   3  2 1
0!  1
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2-Permutation
Permutation is an arrangement of objects in a
specific order. Order matters.

n!
n Pr 
 n  r !
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Example 4-42:
Suppose a business owner has a choice of 5 locations in
which to establish her business. She decides to rank each
location according to certain criteria, such as price of the
store and parking facilities. How many different ways can
she rank the 5 locations?
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Solution :
5!= 5.4.3.2.1= 120
5!
5! 5.4.3.2.1
 
 120
5 p5 
(5  5)! 0!
1
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Example 4-44:
A television news director wishes to use 3 news stories on an
evening show One story will be the lead story, one will be the
second story, and the last will be a closing story. If the director has
a total of 8 stories to choose from, how many possible ways can the
program be set up?.
Solution :
Since there is a lead, second, and closing story, we know that order
matters. We will use permutations.
8!
 336
8 P3 
5!
or
P  8  7  6  336
8 3
3
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3-Combination
 Combination is a grouping of objects. Order
does not matter.
n!
n Cr 
 n  r  !r !
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Example 4-47:
How many combinations of 4 objects are there . Taken 2 at a
time?
Solution :
This is a combination problem , the answer is
4c2
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Example 4-49:
• In a club there are 7 women and 5 men. A committee of 3 women
and 2 men is to be chosen. How many different possibilities are
there?
Solution :
There are not separate roles listed for each committee member,
so order does not matter. We will use combinations.
7!
5!
Women: 7C3 
 35, Men: 5C2 
 10
4!3!
3!2!
There are 35·10=350 different possibilities.
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Permutation
Means arrangement of
things.
combination
Means selection of things
Notes:
1. nPn=n!
2. nCn=1
3.The relation between combination and
permutation
n Pr
n
Cr 
r!
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Probability and
Counting Rules
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Example 4-51:
• A box contains 24 transistors , 4 of which are defective . If 4
are sold at random , find the following probabilities :
a- Exactly 2 are defective .
C2 .20 C2 1140
P(exactly2aredefectives) 

10626
24 C 4
4
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•
b-Non is defective .
C4 4845
P(nodefectives ) 

10626
24 C 4
20
C- All are defective .
C4
1
P(alldefective) 

10626
24 C 4
4
D- At least 1 is defective
P (atleast1defective)  1  P(nodefective)
C4
1615 1927
 1
 1

3542
3542
24 C 4
20
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Example 4-52:
• A store has 6 TV Graphic magazines and 8 Newstime
magazines on the counter . If two customers purchased a
magazine, find the probability that one of each magazine
was purchased.
Solution:
P( 1 TV Graphic and 1 Newstime)

6
C1.8 C1 6.8 48


91 91
14 C 2
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Example 4-53:
• A combination lock consist of the 26 letters of the alphabet .
If a 3- letter combination is needed , find the probability that
the combination will consist of the letters ABC in that order
.The same letter can be used more than once .
Solution :
1
1
P( ABC )  3 
26
17576
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Example 4-54:
• There are 8 married couples in a tennis club . If 1 man and 1
woman are selected at random to plan the summer
tournament , find the probability that they are married to
each other .
Solution :
P(they are married to each other )=
1
1

8.8
64
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
How many different ways can be selecting 4
bananas and 3 apples from 6 bananas and 8
apples?
A)120
B)696
C)840
Anc. C
D)71
• How many different words can we make using the
letters A, B, E and L and repetition is not allowed?
A-25
B-26
C-23
Anc. D
D-24
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 A jar contains 3 red marbles, 7 green marbles and 10
white marbles. If a marble is drawn from the jar at
random, what is the probability that this marble is white?
A)0.5
B)0.4
C)0.3
Anc.A
D)0.8
 Given ten employees, three of them are males . If two
employees are selected at random , what is the
probability that both of them are females?
A)1/15
B)3/50
C)7/15
Anc.C
D)21/50
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 If P(A)=0.6, P(B)=0.3and P(A or B)=0.7, given that
A,B are not mutually exclusive , Find P(A and B)?
A)0.2
B)1
C)1.5
ANC.A
D)0.18
 The events A and B are such that P(A)=0.35 ,
P(B)=0.55 . Find P(A∪B) if A and B are independent
events .
a) 0.19
b) 0.71
c) 0.9
ANC.C
d) 1.02
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 A fair coin is toss and a dice is drown .Find the
probability of getting a head and a number less than
5.
a) 1/3
b) 2/12
 4/3
ANC.A
 6/9
 The number of different ways to arrange four numbers
from 2,4,6,8,7,9 :
a) 360
b) 120
c) 321
ANC.A
d) 432
Note: This PowerPoint is only a summary and your main source should be the book.
Note: This PowerPoint is only a summary and your main source should be the book.