This document discusses counting rules and methods for determining the number of possible outcomes of an event. It begins by explaining that counting rules are needed when the number of possibilities is too large to list completely, such as when tossing coins and dice. The fundamental counting rule states that if one operation can be done in n1 ways and a second in n2 ways, the combined operations can be done in n1×n2 ways. Permutations refer to arrangements that consider order important, while combinations disregard order. Formulas are provided for calculating permutations and combinations. Examples demonstrate applying these rules and formulas to problems involving arranging objects and selecting subsets of items.

1. 1

2. COUNTING RULES
PRESENTED BY
AMJAD ALI CHAUDHRY
2

3. TREE DIAGRAM
• Most often we need to determine the number of ways something can
happen
• The number of possible outcome for an experiment
• The number of ways an event can occur
• The number of ways a certain task can be performed, and so forth
• In some situations we can construct a tree diagram to determine a
complete list.
3

4. 4
TREE
DIAGRAM
A coin and six sided die is tossed simultaneously. How many
possibilities are there?
T
1
2
3
4
5
6
H
1
2
3
4
5
6
Coin
H,1
H,2
H,3
H,4
H,5
H,6
T,1
T,2
T,3
T,4
T,5
T,6
Coin Die Outcome
Amjad Ali

5. WHY LEARN COUNTING RULES ?
5
• We may construct tree diagram to obtain the complete list when
number of possibilities is small
• In most cases, the number of possibilities is too large to permit a
complete listing.
• Thus we need to develop techniques/ rules that enable us to determine
the number of possible ways something can happen.
• For instance, in an experiment of tossing one coin and two dice
simultaneously, the number of possibilities is too large to construct a
tree diagram. In such situations need some counting rules to determine
the number of all possible outcomes
Amjad Ali

6. 6
FUNDAMENTAL COUNTING RULE
This rule states that if an operation can be performed in n1 ways, and if for each of these
a second operation can be performed in n2 ways, then the two operations can be
performed together in n1 × n2 ways.
EXAMPLE
A coin and six sided die is tossed simultaneously. How many possibilities are there?
A coin has two possible outcomes: n1 = 2
A die has six possible outcomes: n2 = 6
Total number of possibilities = n1 × n2
= 6 × 2 = 12.
Amjad Ali

7. 7
EXTENSION OF FUNDAMENTAL COUNTING RULE
If an operation can be performed in n1 ways and for each of these a second operation
can be performed in n2 ways, and for each of the first two a third operation can be
performed in n3 ways and so forth, then the sequence of k operations can be performed
in n1 × n2 × n3 × …… × nk ways.
EXAMPLE
A coin and two six sided dice are tossed simultaneously. How many possibilities are
there?.
Total number of possibilities are: n1 × n2 × n3 = 2 × 6 × 6 = 72
EXAMPLE
How many outcomes are possible when three fair dice are rolled?
Total possible outcomes are: n1 × n2 × n3 = 6 × 6 × 6 = 216
Self Test
How many outcomes are possible when 4 coins are tossed?
Amjad Ali

8. 8
EXAMPLE
A cafeteria offers a choice of two soups, five sandwiches, three desserts, and three
drinks. How many different lunches, each consisting of a soup, a sandwich, a dessert,
and drink are possible?
A soup can be chosen in n1 = 2 different ways.
For each of these a sandwich can be chosen in n2 = 5 different ways
For each soup- sandwich combination, a dessert can be chosen in n3 = 3 different ways
A drink in n4 = 3 different ways
Total number of different lunches of a soup, a sandwich, a dessert, and a drink is:
n1 × n2 × n3 × n4 = 2 × 5 × 3 × 3 = 90
Amjad Ali

9. 9
FACTORIAL
Let n be a positive number, then the factorial of n is denoted by n! and is defined as the
product of all integers from n to 1. In other words,
n! = n (n – 1) (n – 2) (n – 3)………3. 2. 1
To make the factorial notation consistent, we make the convention 0! = 1
Other examples of factorial are:
1! = 1 4! = 4×3 ×2 ×1 = 24
2! = 2 ×1 = 2 5! = 5 ×4 ×3 ×2 ×1 =120
3! = 3 ×2 ×1 = 6 6! = 6 ×5 ×4 ×3 ×2 ×1 = 720
One property of factorial notation is that it can be stopped at any point by using the
exclamation point.
n! = n (n – 1)!
= n (n – 1) (n – 2)!
= n (n – 1) (n – 2) (n – 3)! and so on
Amjad Ali

10. 10
EXAMPLE
Suppose that a student has three books on statistics named A, B, and C to arrange on a
shelf of a bookcase. In how many ways the three books be arranged?
A
B C
C B
A B C
A C B
B
A C B A C
C A B C A
C
A
B
B
A C B A
C A B
Amjad Ali

11. 11
Suppose that a student has 4 books named A, B, C, D to arrange on a shelf of a
bookcase. How many possibilities are there?
The answer is 4! = 24.
Let us construct the Tree Diagram
It is clear from tree diagram that these books can be arranged on the shelf in 6 possible
ways
The first book can be chosen and placed in n1 = 3 different ways
Now apply the COUNTING RULE for the same example
After placing the first, the second book can be chosen and placed in n2 = 2 different
ways and the third in n3 = 1 way.
Hence the total number of ways is equal to
n1 × n2 × n3 = 3 × 2 × 1 = 6
Again we observe that there are 6 ordered arrangement of 3 books. This agrees with
the result that 3! = 6
Amjad Ali

12. 12 Amjad Ali
D
B
C
B
C
D
C
B
D
B
D
C
D
A C
B
A B C D
A B D C
A C B D
A C D B
A D B C
A D C B
D
A
C
A
C
D
C
A
D
A
D
C
D
B C
A
B A C D
B A D C
B C A D
B C D A
B D A C
B D C A

13. 13
D
A
B
A
B
D
B
A
D
A
D
B
D
C B
A
C A B D
C A D B
C B A D
C B D A
C D A B
C D B A
C
A
B
A
B
C
B
A
C
A
C
B
C
D B
A
D A B C
D A C B
D B A C
D B C A
D C A B
D C B A
Amjad Ali

14. 14
Apply Extension of Fundamental Counting Rule
The first book can be chosen and placed in n1 = 4 different ways.
The second book can be chosen and placed in n2 = 3 different ways.
The third book can be chosen and placed in n3 = 2 different ways.
and the fourth in n4 = 1 way.
Hence the total number of ways is equal to
n1 × n2 × n3 × n4 = 4 × 3 × 2 × 1 = 24
This agrees with the result that 4! = 24
What should we do to find number of possibilities in such situations? Do not think
much, use n!
Amjad Ali

15. 15
NOTE:
In previous example we selected all the objects. Sometimes, however, we want to
select only some of the n objects. Assuming that we have n different objects available
and we want to select r of them, how many different ordered arrangements are
possible? To answer this question considers the following example
EXAMPLE
Suppose that a student has four books on statistics named A, B, C, and D to arrange on
a shelf, but there is room for only two books on the shelf, how many ways can these
books be arranged on the shelf?
Amjad Ali

16. 16
A C
B
D
A B
A C
A D
B C
A
D
B A
B C
B D
Amjad Ali

17. 17
C B
A
D
C A
C B
C D
D B
A
C
D A
B C
D C
Amjad Ali

18. 18
We can conduct that there are 12 ordered arrangements which are AB, AC, AD, BA, BC,
BD, CA, CB, CD, DA, DB, and DC.
This again could be obtained by using the fundamental counting rule.
The first book can be chosen and placed in 4 ways, and, once selected, the second book
can be chosen and placed in three ways. Hence the total number of ways = 4 ×3 = 12
(These 12 ordered arrangements are 12 permutations)
Self Test
Consider an example of placing 3 of the 5 books named A, B, C, D, and E. Think that
how many possibilities are there?
Do you think that answer is 60?
Amjad Ali

19. 19
PERMUTATION
A permutations of a number of objects is an arrangement of these objects in a specific
order
EXAMPLE
Suppose that six books A, B, C, D, E and F to arrange on a shelf, but there is room for
only two books on the shelf, how many possibilities (permutations) are there?
Permutations are
A B B A C A D A E A F A
A C B C C B D B E B F B
A D B D C D D C E C F C
A E B E C E D E E D F D
A F B F C F D F E F F E
Note that that there are 30 possible permutations
Amjad Ali

20. 20
For previous example, observe that
first space can be filled in n1 = 6 ways,
and second space can be filled in n2 = 5 ways.
Using fundamental counting rule, total possibilities (permutations) are: n1 × n2 = 6 × 5
= 30
Note: 6 × 5 could written as
In general, the number of permutations or ordered arrangements of r objects taken
from n distinct objects is written as nPr, is given by
NOTE:
!
2)
-
(6
!
6
!
4
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6
1
2
3
4
1
2
3
4
5
6
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

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
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


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!
r)
-
(n
!
n
Pr
n

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n
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n)
-
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n
Pn
n


Amjad Ali

21. 21
PERMUTATION RULE
The number of permutations of n distinct objects, taking r objects at a time, denoted by
nPr, is given by
EXAMPLE
A club consists of 15 members. In how many ways can be three officers: president, vice
president, and secretary treasurer, be chosen?
The order, in which 3 officers are to be chosen, is of significance, therefore
Number of permutations of n distinct objects taken all at a time is equal to n!
!
r)
-
(n
!
n
Pr
n

2730
!
3)
-
(15
!
15
P3
15


Amjad Ali

22. 22
COMBINATIONS
We learned that a permutation is an arrangement of n objects in a specific order.
However, in some situations, order of selection is not important but only the objects
that are chosen.
For example, we wish to select a two-person committee from a group of six students.
Here the order of choice is unimportant because all six students will be equal members
of the committee. Suppose the six students are named as A, B, C, D, E, and F, then the
possible selections are
Combinations
A B B C C D D E E F
A C B D C E D F
A D B E C F
A E B F
A F
These 15 selections are called fifteen combinations. Order of selection in this case is of
no significance. Thus we write BA or AB both these arrangements represent only one
combination Amjad Ali

23. 23
EXAMPLE: Given the letters A, B, C, D, E, and F. List the permutations and combinations
for selecting two letters
COMBINATION
A selection of distinct objects without regard to order is called a combination.
Permutations Combinations
AB
AC
AD
AE
AF
BA
BC
BD
BE
BF
CA
CB
CD
CE
CF
DA
DB
DC
DE
DF
EA
EB
EC
ED
EF
FA
FB
FC
FD
FE
AB
AC
AD
AE
AF
BC
BD
BE
BF
CD
CE
CF
DE
DF
EF
We see that each combination gives rise to 2! Permutations, or, vice versa, to each 2!
permutations these corresponds one combination.
Here TOTAL COMBINATIONS (2!) = TOTAL PERMUTATIONS
If the number of combinations of two letters from 6 letters is denoted by 6C2 then
2
6
2
6
P
)
!
2
(
C 
r)!
-
(n
!
r
!
n
C
general
in
and
2)!
-
(6
!
2
!
6
!
2
P
C r
n
2
6
2
6



Amjad Ali

24. 24
COMBINATION RULE
The number of possible combinations of n distinct objects taken r at a time, denoted by
nCr, is given by nCr =
n!
r! (n – r)!
EXAMPLE
An English department at a university has 15 faculty members. Three of the faculty
members will be randomly selected to represent the department on a committee. How
many different selections are possible?
In forming the committee, order of selection is irrelevant. Therefore, we want the
number of combinations of 15 faculty members when 3 are selected. We get
15C3 = 455
Amjad Ali

25. 25
RED BLACK TOTAL
4 2 6
Total outcomes are = 21
EXAMPLE
Two balls are drawn together at random from a bag containing 4 red and 2 black balls.
How many possible outcomes are there?
1 2 1 3 1 4 1 1 1 2
2 2 2 1 2 2
3 3 1 3 2
4 1 4 2
1
3 4
4
2
Using Combination Rule
6C2 = 15
Amjad Ali

26. THANK
YOU
Counting
Rules
Amjad Ali Chaudhry
26