Probability module 1

Module 1 Counting Techniques
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PROBABILITY and STATISTICS
COUNTING TECHNIQUES
Module 1
RICHARD B. PAULINO
Ilocos Norte Regional School of Fisheries

Counting Techniques
What this module is all about?
PROBABILITY
PROBABILITY connotes the “chance” or the “likelihood” that something will happen or occur. If
one has to make precision regarding the outcome of a certain activity, then the possible outcomes should
first be identified. Identification requires the knowledge of counting.
In this module, we will discuss the most commonly used methods of counting. The problem of
counting number of ways arranging a set of objects or the number of alternative ways of doing particular
task has been of interest to human being for generations.
This module provides you an adequate discussion of the different methods used in counting. In
counting, the simplest method is done by enumerating all the possible outcomes. This method, however
is too laborious and hence not an efficient method. The number of different possible outcomes may
sometimes become extremely large so that the usual counting procedure becomes impractical. In this
module I will consider the most commonly used methods of counting – the fundamental principle of
counting, permutation and combination.
Your knowledge about these methods shall be your foundation in understanding other concepts in
probability.
The following topics are contained in this module
 Lesson 1. The Fundamental Principle of Counting
 The Tree Diagram
 Multiplication Rule
 Lesson 2 Permutation
 Permutation of n distinct objects.
 Permutation of n distinct objects taken r at a time.
 Circular Permutations
 Permutation with things that are alike
 Lesson 3. Combination
ENJOY READING THIS MODULE!!!
What you are expected to learn.
After thorough understanding of all the topics and had performed all the activities in this module,
you are expected to:
 Apply the tree diagram as a counting method;
 State the multiplication rule;
 Apply the multiplication rule in predicting outcomes;
 Define permutation;
 Apply permutation in counting the arrangement of n distinct objects;
 Calculate the permutations with things that are alike
 Define combination and
 Apply combinations as a counting technique

How you are going to learn?
This is the icon that you will find before some chunks of text in
the following pages. It tells you to study carefully the
concepts, principles, processes, etc. discuss in the text.
This is the icon signals a checkpoint. At every checkpoint, you
will find a question or questions to answer.
This is the icon introduces a list of important ideas to
remember. Read it carefully and store it in your memory.
This icon signals a self- test to determine how well you had
achieved the objectives of the module. Study the module
carefully and you will perform quite well in the self test.
This icon tells you of an assignment to perform. The quality of
your output from this assignment will determine the extent of
what you learned in this module.

LESSON 1
In the study of “how many” or “what is possible” there are essentially two types of procedures.
Firstly, a complete listing of all possibilities in a given experiment and secondly, determining the number
of ways a particular task can be done.
Consider the numbers 1, 2, and 3. Suppose we want to determine the total two-digit numbers that
can be formed if these numbers are combined. First, let us assume that no digit is to be repeated. Thus,
the possible two-digit numbers that can be formed can be enumerated as follows:
12 21 31
13 23 32
One way of solving this is to list all possible combinations of units and counting them. This is the use of a
TREE DIAGRAM as illustrated below.
First Digit Second Digit Outcome
2 12
1
3 13
1 21
2
3 23
1 31
3
2 32
Notice we were able to exhaust all the possibilities through enumeration in this example, we have
six possible two-digit numbers. Now, what if the digits can be repeated? If repetition is allowed, then the
possible two-digit numbers are as follows.
11 21 31
12 22 32
13 23 33
Let us try to illustrate this using the tree diagram.
Using the tree diagram, we have
First Digit Second Digit Outcome
1 11
1 2 12
3 13
1 21
2 2 22
3 23
1 31
3 2 32
3 33
Hence, we have 9 possible outcomes.
FUNDAMENTAL PRINCIPLE OF COUNTING
A TREE DIAGRAM provides a clear and orderly way of listing all possible
combinations by counting the various paths along the branches.

The following examples will help you understand the process of counting using the tree diagram.
Example 1
In how many ways can a working student go home from school via his office if there are three
routes from his school (S) to his office (O) and two routes from his office to his home (H)?
Solution:
We list all possible combinations of routes and counting them. A tree diagram will help us to do
this
O H WAYS
1 A-1
A
2 A-2
1 B-1
S B
2 B-2
1 C-1
C
2 C-2
The diagram showed that there are 6 ways of going home from school via his office : A-1, A-2, B-1,
B-2, C-1, C2.
Example 2
If a coin and die are tossed simultaneously, how many events are possible?
Solution:
One toss of a coin will result in one of two outcomes, head or tail, while one
toss of a die will result in one of six possible outcomes, 1, 2, 3, 4,5, or 6 dots showing on the top face.
Using the three diagram, there are 12 possible outcomes that would occur as illustrated.
COIN DIE OUTCOME
H
1 H-1
2 H-2
3 H-3
4 H-4
5 H-5
6 H-6
T
1 T-1
2 T-2
3 T-3
4 T-4
5 T-5
6 T-6
Let us now consider the total three-digit numbers that can be formed from the numbers 1, 2, 3, 4
and 5. Clearly, the number of possibilities is now large
The TREE DIAGRAM as a method of enumeration can always be
applied if the number of possibilities is at manageable level.

Thus, we have to use another counting technique – the Fundamental Principle of Counting or
the Multiplication Rule
Therefore, in example 1, if there are 3 routes from school to the office and 2 routes from office to
his home, the number of ways in which the student can reach home from school via his office is
3 x 2 = 6 ways.
Let us apply the multiplication rule to the following examples.
Example 3:
How many two digit number can be formed from 3,4,5,6 and 7if no digit is repeated?
Solution:
To form numerals of two-digits, we have to fill up the tens and unit digits. For the tens places,
there are five choices and since no digit is repeated, the units places can be filled up with anyone of the
remaining four digits. Therefore, there are 5 x 4 = 20 two-digit number that can be formed as shown.
34 35 36 37
43 45 46 47
53 54 56 57
63 64 65 67
73 74 75 76
The principle can obviously be extended to cases with more than two activities.
Example 4
How many three – digit numbers can be formed from the digits 1, 2, 3, 4 and 5 if any of the digits
can be repeated?
Solution:
We let n1 be the number of ways of filling the hundreds place. Let n2 be the number of ways of
filling the hundreds place. Let n2 be the number of ways of filling the tens place and n3 be the number of
ways of filling the units place.
The hundreds place can be filled in 5 ways. Since repetition of numbers is allowed then the tens
and the units can both be filled in 5 ways. Thus, n1 = 5, n2 = 5 and n3 = 5. Then by the multiplication rule,
we have
n1 x n2 x n3 = 5 x 5 x 5 = 125 ways
When an operation consists of more than 2 steps made in succession, the
number of ways this operation can be completed is
n1 x n2 x n3 x ......x nk
where n1 , n2, n3, and nk , are the first , second, third and k steps respectively.
The MULTIPLICATION RULE states that if the first activity can be
done n1 ways and after it has been done, the second activity can
be done in n2 ways, then the total number of ways in which the
two activities can be done is equal to n1 x n2.

Example 5
How many ways can the Miss Universe, the first runner and second runner-up be chosen from 10
finalists?
Solution:
Let n1 be the number of ways of choosing the Miss Universe Let n2 be the number of ways of
choosing the first runner-up and n3 be the number of choosing second runner-up.
The Miss Universe can be chosen in 10 possible ways. After choosing the Miss Universe, from the
9 finalists left, the first runner-up can be selected in 9 ways. Finally, after choosing the Miss Universe and
the first runner-up, the second runner-up can be selected from the remaining 8 finalist in 8 ways. Hence,
by the fundamental principle of counting, we have
n1 x n2 x n3 = 10 x 9 x 8 = 720
Do you want another illustrative problem? Okey, here it is.
Example 6
Consider the number 2, 3, 5, and 7. If repetition is not allowed , how many three-digit numbers
can be formed such that
a. they are all odd
b. they are all even
c. they are greater than 500 ?
Solution:
Let n1 = number of ways of filling the hundreds place
n2 = number of ways of filling the tens place
n3 = number of ways of filling the units place
a. The units place can be filled by the numbers 3, 5 or 7 to make the required numbers odd. Hence,
we can say that n3 = 3. Since repetition is not allowed, then one of the numbers 3, 5 or 7 can no
longer be used in the tens place. However, number 2 can now be included in the tens place. Thus,
n2 = 3. Of the three values used in the tens place, one value can no longer be used in the hundred
place. Thus n1 = 2. Then by the multiplication rule, we have
n1 x n2 x n3 = 2 x 3 x 3 = 18 ways
To verify, the required numbers are
235 237 253 257 273 275
325 327 357 375 523 527
573 537 723 725 735 753
b. Since the requirements is the set of three-digit even numbers, then the first logical step is done by
filling units place by the number 2. Thus, we can say that n3 = 1. It was mentioned that repetition
is not allowed, so number 2 can no longer be used in the tens place. This leaves us in three ways:
3, 5 or 7. Thus the value of n2 is 3. In the case of hundreds place, after choosing the tens place,
there shall still be two numbers available and hence n1 = 2. By using the fundamental principle of
counting, we have
n1 x n2 x n3 = 2 x 3 x 1 = 6 ways
The six three-digit number are shown below.
352 532 732
372 572 752

c. The condition that the numbers are greater than 500 requires us not to place numbers 2 and 3in
the hundreds place but can only use the numbers 5 or 7. Hence we can say that n1 = 2. The tens
place can be using the numbers 2 and 3 and only of the numbers 5 or 7. Thus the value of n2 = 3
The units place therefore can be place in two different ways or n3 = 2. Using the multiplication
rule, we have,
n1 x n2 x n3 = 2 x 3 x 2 = 12 ways
The twelve three-digit number are shown below
SELF-TEST I
Solve the following problems
1. Show by tree diagram how many meals consisting of fruit, a sandwich and a drink are possible if
one can select from 3 kinds of fruits (mango, banana, papaya) 3 kinds of sandwiches (egg, cheese,
tuna) and 3 kinds of drinks ( coffee, milk, juice).
2. Considers the number 0, 3, 5, 6 and 8.
How many two-digit numbers can be formed such that
a. they are all even?
b. they are all odd?
c. they are greater than 50?
3. A college has 3 entrance gates and 2 exit gates. In how many ways can a student enter then leave
the building ?
4. How many possible outcomes are there if a pair of dice is rolled? Use the tree diagram to illustrate
this?
5. How many nonsense words of 3 letters can be formed from the letters c, d, e, f and g if
a. repetition of letters is allowed?
b. repetition of letters is not allowed?
523 572 732
527 573 735
532 723 752
537 725 753
You learned in the previous example that the number of possible outcomes can be determine by
enumeration or by by usin the TREE DIAGRAM or by the MULTIPLICATION RULE

LESSON 2
Frequently, we are interested in a set of possible outcomes that contains as elements all possible
orders or arrangements of a group of objects. For example, we want to know how many different
arrangements are possible for sitting 6 people around a table, or we may ask how many different orders
are possible for drawing 2 lottery tickets from a total of 20.
The different arrangements are called Permutations.
Permutation of n distinct objects.
Consider the three letters a, b and c. The possible arrangements are abc, bac, bca, cab and cba.
Thus we see that there are 6 distinct arrangements. Using the multiplication rule, we arrive at answer 6
without actually listing the different orders. There are n1 = 3 choices for the first , then n2 = 2 for the
second . Leaving only n3 = 1 choice for the last position giving a total of
n3 x n2 x n1 = 3 x 2 x 1 = 6 permutations
In general, n distinct objects can be arranged in n(n-1)(n-2)(n-3)...(3)(2)(1) ways.
We represent this product by the symbol n!, which is read “n factorial”. Three objects can be
arranged in 3! = (3)(2)(1) = 6 ways. By definition 1! = 1 and 0! = 1.
It should be noted that 5! ≠ 5 since 5! = 120. In the same manner 3! + 4! ≠ 7!. The left hand side is
equal to
3! + 4! = (3)(2)(1) + (4)(3)(2)(1)
= 6 + 24 = 30
3! + 4! ≠ 7!
The number of permutations of the four letters a,b,c and d will be 4! = 24
PERMUTATION
A PERMUTATION is an arrangement of all or part of a set of objects
with reference to order
The number of permutations of n distinct objects taken all at a time is n!
nPn = n!

Example 1
In how many ways can 5 different books be arranged on a shelf?
Solution:
5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120 ways
Example 2
Find the number of ways in which 6 runners be arranged at the starting line of a race?
Solution:
6P6 = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 ways
Example 3
Suppose 4 different mathematics books and 5 different physics books shall be arranged in a shelf.
In how many ways can such books be arranged if the books of the same subject shall be placed side by
side?
Solution:
 The initial step is to arrange the set of math books as one object and the set of physics
books as another object . This can be done in 2P2 ways.
 The second step is to determine the number of permutation of four distinct math
books. This can be done in 4P4 ways
 The next step is to determine the total number of permutation of 5 distinct physics
books. This can be done in 5P5 ways.
 Then, finally we apply the multiplication rule. Thus
2P2 x 4P4 x 5P5 = 2! x 4! x 5!
= 2 x 24 x 120
= 5,760 ways
The following examples will help
you understand the concept.
Very good! I think you are exhausted now . Relax
for a while. You may take in something to refresh
yourself. Be ready for the next topic.

Permutation of n distinct objects taken r at a time.
Let us now consider the number of permutations that are possible by taking the four letters a, b, c
and d two at a time. These would be ab, ac, ba, ca, ad, ad, bc, cb, db, bd, cd, dc. Using the multiplication
rule, there are two positions to fill with n1 = 4 choices for the first and then n2 = 3 choices for the second
for a total of
n1 x n2 = 4 x 3 = 12 permutations
In general, n distinct objects taken r at a time can be arranged in
n(n-1)(n-2)(n-3)...(n-r+1) ways.
We represent this by the symbol
nPr =
𝒏!
𝒏−𝒓 !
To illustrate, suppose he value of n=7 and r=3. Then the permutation of 7 objects taken 3 at a time
is equal to
nPr =
𝑛!
𝑛−𝑟 !
=
7!
7−3 !
=
7!
4!
=
7𝑥6𝑥5𝑥4𝑥3𝑥2𝑥1
4𝑥3𝑥2𝑥1
= 210 ways
Example 4
In how many ways can a president, a vice president, a secretary and a treasurer be elected from a
class with 39 students?
Solution:
Given 39 students, we are going to fill 4 distinct positions. Hence, we can say that n = 39 and r = 4.
Using the equation, we have
nPr =
𝒏!
𝒏−𝒓 !
=39P4=
𝟑𝟗!
𝟑𝟗−𝟒 !
=
𝟑𝟗!
𝟑𝟓!
=
𝟑𝟗𝒙𝟑𝟖𝒙𝟑𝟕𝒙𝟑𝟔𝒙𝟑𝟓!
𝟑𝟓!
= 1,974,024 ways
Example 5
In how many ways can the first prize, the second prize and the third prize be awarded to ten
entries in an essay contest?
Solution:
Given 10 entries, we are going to award 3 prizes. Hence, n = 10 and r = 3. Using the equation we
have
nPr =
𝑛!
𝑛−𝑟 !
=
10!
10−3 !
=
10!
7!
=
10𝑥9𝑥8𝑥7𝑥6𝑥5𝑥4𝑥3𝑥2𝑥1
7𝑥6𝑥5𝑥4𝑥3𝑥2𝑥1
= 720 ways
The number of permutations of n distinct objects taken r at a time is
nPr =
𝒏!
𝒏−𝒓 !
The following examples will help you understand the concept of
permutation of n objects taken r at a time.

Checkpoint
1. In how many ways can 8 students be arranged in a row of 8 chairs?
2. Find the number of ways in which 6 teachers can be assigned to 4 sections of math class if no
teacher is assigned to more than one section.
3. How many ways can 5 men and 4 women be seated in a row if:
a. they can seat anywhere?
b. the women alternate with the men?
Compare your answer below
Answer: 1.40,320 3. a. 362,880
2. 120 b. 2,880
CIRCULAR PERMUTATIONS
Permutations that occur by arranging objects in a circle are called circular permutations. If the
objects are made to move either in a clockwise or counter clockwise direction without changing the order
or sequence between at least two of them, the resulting arrangement is not considered a different
permutation from the original. Hence, circular permutation of n objects are obtained if one object is
considered fixed in one position and the others are arranged in n -1 distinct arrangements. Consider the
following diagrams where the six different permutations of four objects arranged in a circle are
demonstrated.
Remember
Solve the following problems to test if you understood the two concepts
on permutation that you have just finished studying. Apply the
knowledge you just gained.
Did you get the answer right? That’s a very good job!
Congratulations!
If you did not, check where your mistake is.
Relax!
The number of permutations of n distinct objects arranged in a circle is (n-1)!
This is denoted by 𝑷∘ = (n-1)!

Suppose we want to determine the number of ways of arranging 7 persons in a round table. Then,
by definition, we have
𝑷∘ = (n-1)!
= (7-1)!
=6!
= 6 x 5 x 4 x 3 x 2 x 1 =720 ways
Example 6
In how many ways can 8 colored beads be made into bracelet?
Solution:
From the given condition, we can say that this is a circular permutation. Since n = 8, then
by definition we have
𝑷∘ = (n-1)!
= (8-1)!
=7!
=7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040 ways
Example 7
In how many ways can 6 persons be seated around a table with 6 chairs if;
a. they can seat anywhere?
b. Two individuals wanted to seat side by side?
c. Two particular persons must not seat next to each other?
Solution:
a. By definition, for n = 6 we have
𝑷∘ = (n-1)!
= (6-1)!= 5! = 5 x 4 x 3 x 2 x 1 =120 ways
b. We first consider two individuals who wanted to be seated side by side as one person, we thus
making n = 5. The number of circular permutation of these 5 individuals is (5-1)! = 4! = 24.
Then consider the permutation of the two persons treated as one. They can seat together in
2P2 = 2! = 2 ways. Hence, by the fundamental principle of counting, we have
𝑷∘ = 24 x 2 = 48 ways
c. The total ways for the 6 persons to seat anywhere around the table is 120 ways. There are 48
ways of arranging them with two particular persons seated side by side. Then, the number of
ways of arranging the 6 persons if two particular persons must not seat next to each other is
the difference. Hence, we have
𝑷∘ = 120 – 48 = 72 ways
Here some illustrative examples that will help explain further the concept of
circular permutation. Study them carefully.

PERMUTATION WITH THINGS THAT ARE ALIKE
In the preceding section, we established the assumption that n objects are distinct. There are
some instances when some objects cannot be distinguished from one another since such objects are alike.
Consider for example the letters from the word MOMMY. If 3 M’s shall be distinguished from each other,
then the total number of permutations that can be formed from the letters of the given word is 5P5 or
120 ways. Such is not the case since 3 M’s are alike. Thus we have to develop a formula that shall suit this
situation.
Remember
So, if we want to determine the number of permutations that can be formed using the word
MOMMY, we have the following solutions.
The value of n is 5 because there are 5 letters to be arranged. We let r1 be the number of M’s, r2
be number of O’s and r3 be the number of Y’s. We can say that r1=3, r2=1 and r3=1. Using the equation,
we have
5P5 =
5!
3!𝑥1!𝑥1!
= 20 permutations
Example 8
Determine the possible permutations of the word MISSISSIPPI.
Solution:
The value of n = 11. Let r1 the number of M’s,
r2 be the number of I’s,
r3 be the number of P’s.
By inspection, we can say that r1 = 1, r2=4 , r3=4 and r4=2,
using the equation we have
11P11 =
11!
1!𝑥4!𝑥4!𝑥2!
= 34,650 ways
Example 9
In how many ways can 3 algebra, 5 calculus and 7 trigonometry books be arranged in a shelf?
Solution:
From the given conditions we can say that n = 3 + 5 + 7 = 15, r1 = 3, r2=5 , and r3=7.
Using the equation, we have
15P15 =
15!
3!𝑥5!𝑥7!
= 360,360 ways
The number of permutations of n objects taken altogether, where r1
are of one kind , r2 are of the other kind and so on is given by
nPn =
𝒏!
𝒓 𝟏!𝒙𝒓 𝟐!𝒙…𝒙𝒓 𝒌!
Here are some illustrative examples that will help you
understand the concept.

Checkpoint
1. In how many ways can 8 persons ride in a merry-go-round?
2. Determine the number of distinct permutations that can be formed from the word STATISTICS?
Answers: 1. 5,040 2. 50,400
SELF-TEST 2
1. In how many ways can 9 passengers be seated in a bus if there are only 5 seats available?
2. In how many ways can 5 different shirts be arranged in a cabinet?
3. Five red marbles, two white marbles and three blue marbles are arranged in a row. If all the
marbles of the same color are not distinguishable from each other, how many different
arrangements are possible?
4. In how many ways can 4 boys and 4 girls be seated in a row if
a. they can seat anywhere
b. the boys and the girls are to be seated alternately?
5. Determine the number of distinct permutations that can be formed from the word PHILIPPINES.
6. In how many ways can 5 precious stones be set around a gold bracelet?
Here are some problems which can serve as your checkpoint if you
understood the last two concepts discussed. Try to solve them and
compare your answers below.
Did you get the answer right? That’s a very good job!
Congratulations!
If you did not, check where your mistake is.
Relax!
That’s a very good job!
Congratulations!
You deserve points for that. You may now proceed to
the next lesson.

LESSON 3
In permutations, order is very important in determining the number of possible arrangements of
the given objects. In some cases, however, no importance is attached to the order with which objects are
arranged. In many situations, we are interested in determining the number of ways of selecting r objects
from a total of n objects without regard to order. Such situation calls for the study of combinations.
A COMBINATION is a group of objects where the composition of the group, but not the order is
important. For example, given the letters a, b, c the number of two-letter combinations that may be
formed are only 3. These are ac, ab and bc. The number of two-letter permutations here is six; ab, ac, ca,
cb, ba and bc. In combinations, the order of the objects is not important such that ab and ba are merely
counted as one and not two different combinations.
Remember
This idea of combination is usually applied to problems about groups, committees or collections
where order of the elements is not important.
Consider the number 1, 2, 3, 4 and 5. Suppose we take two elements from this group of five
numbers. If order is important and no repetition is allowed, then the total number of elements that can be
formed is 20, computed as follows;
5P2 =
𝑛!
𝑛−𝑟 !
=
5!
5−2 !
= 20
The 20 two-digit numbers are shown below:
12 21 31 41 51
13 23 32 42 52
14 24 34 43 53
15 25 35 45 54
Notice that since order is important, the numbers 12 and 21 are different numbers. Similarly, the
numbers 42 and 24 are treated as different numbers. In the case of combination, the numbers 12 and 21
represent the same grouping since the order in which the objects are taken is not important. Thus if in
permutations, we have 20 possibilities, in combination we have only 10 as shown below.
nCr or C(n,r)=
𝑛!
𝑟! 𝑛−𝑟 !
5C2 or C(5,2)=
5!
2! 5−2 !
=
5!
2! 3 !
5𝑥4𝑥3𝑥2𝑥1
2𝑥1 3𝑥2𝑥1
=10
COMBINATIONS
The number of combinations of n objects taken r at a time is
nCr or C(n,r)=
𝒏!
𝒓! 𝒏−𝒓 !

Remember
Example 1
In how many ways can a committee of 3 members be chosen from a group with 6 members?
Solution:
Grouping for committees is treated as a combination problem since only one position is
being filled, that is, committee membership. Thus, we have a situation of having 6 objects taken 3 at a
time. Using equation, we have
6C3 or C(6,3)=
6!
3! 6−3 !
=
6!
3! 3 !
6𝑥5𝑥4𝑥3𝑥2𝑥1
3𝑥2𝑥1 3𝑥2𝑥1
=20 ways
Example 2
In how many ways can a student answer 5 out of 8 questions?
Solution:
The student can answer 5 out of 8 questions in 6C5 ways. Hence
8C5 or C(8,5)=
8!
5! 8−5 !
=
8!
5! 3 !
8𝑥7𝑥6𝑥5𝑥4𝑥3𝑥2𝑥1
5𝑥4𝑥3𝑥2𝑥1 3𝑥2𝑥1
=56 ways
Example 3
In how many ways can a student answer 5 out of 8 questions if he is required to answer 3 of the 4
questions?
Solution:
Since 3 of the first questions must be answered, then this may be done in 4C3 ways. Having now
answered 3 questions, then he must answer 2 of the remaining 4 questions. This can be done in 4C2 . Using
the Fundamental Principle of Counting, the total ways in which the student can answer the examination is
given by
4C3 x 4C2 =
4!
3! 4−3 !
X
4!
2! 4−2 !
=
4!
3! 1 !
X
4!
2! 2 !
=
4𝑥3𝑥2𝑥1
3𝑥2𝑥1 1
x
4𝑥3𝑥2𝑥1
2𝑥1 2𝑥1
= 4 x 6
= 24 ways
In combination, the order in which the object are taken is not important
To understand combinations better, let us solve together
the following example problems.

Example 4
A box contains 7 red and 6 green balls. In how many ways can 2 balls be drawn if
a. order is not important?
b. they are both green?
c. 1 is red and 1 is green?
Solutions:
a. There is no condition what ball must be drawn from the box. Hence, we are only
required to determine the number of ways of drawing 2 balls from a box with 13 balls.
This can be done in 13C2 ways. Thus,
13C2 =
13!
2! 13−2 !
=
13!
2! 11 !
= 78 ways
b. The condition requires that two balls when drawn must be both green. This can be
done in 6C2 . Since only two balls shall be drawn, it follows that no red ball shall be
taken which be done in 7C0 way. Therefore, by the Fundamental Principle of Counting,
we have
6C2 x 7C0 =
6!
2! 6−2 !
X
7!
0! 7 !
= 15 ways
c. One red ball can be taken from 7 red balls in 7C1 ways. One green ball can also be
taken from 6 green balls in 6C1 ways. Therefore, by the Fundamental Principle of
Counting, we have
7C1 x 6C1 =
7!
1! 7−1 !
X
6!
1! 6−1 !
=
7!
1! 6 !
X
6!
1! 5 !
= 42 ways
Example 5
Seven male and 10 female math teachers applied for position at a certain university. In how many
ways can 6 teachers be accepted if 3 of the teachers are females?
Solution:
The3 female teachers can be selected in 10C3 ways, while the remaining 3 teachers can be
taken from the 7 male teachers in 7C3 ways. Using the fundamental principle of counting, we have
10C3 x 7C3 =
10!
3! 10−3 !
X
7!
3! 7−3 !
=
10!
3! 7 !
X
7!
3! 4 !
= 4,200 ways
Example 6
How many combinations of three-digit numbers are possible with the digits 3, 5, 1?
Example 7
In how many ways can a committee of 4 members be formed from 5 couples if husband and wife
are not to be in the same committee?
Answers: 1 , 25
Alright!
Try to solve the problems below to see if you understood the
concept.
Compare your answers below.
VERY GOOD!
JOB WELL DONE!
I suppose you are tired now. Take some break. After the break, go
over the lesson and get ready for the self-test.

SELF-TEST 3
1. In how many ways can five patients be accommodated in a hospital ward with 10 beds?
2. In how many ways can a baseball tem of 9 players be formed from a group of 18 players, if
only 3 players can be catchers, only 6 can be pitchers, and the rest can only take one of the
other positions.
3. Out of 5 chemist and 7 physicists, a committee consisting of 2 chemists and 3 physicists is to
be formed. In how many ways can this be done if
a. Any chemist and any physicist can be included?
b. One particular physicist must be on the committee?
c. Two particular chemists cannot be on the committee?
4. From 4 red, 5 green and 6 yellow apples, how many selections of 9 apples are possible if 3 of
each color are to be selected?
Assignment
1. From the ten digits 0,12,3,4,5,6,7,8,9, how many four-digit numbers can be formed if
a. repetitions are allowed?
b. repetitions are not allowed
c. the last digit is zero and no repetition of digits?
d. the tens digit should be odd?
2. Using the tree diagram, how many possible outcomes can result if you toss 4 coins
simultaneously?
3. How many permutations are possible with the letters of the word BETWEENNESS?
4. In how many ways can 5 different trees be planted in a circle?
5. How many different ways can 3 red, 4 yellow and 2 blue bulbs be arranged in a string of
Christmas tree lights with 9 sockets?
6. Nine people are going on a trip in 3 cars that will hold 2, 4, and 5 passengers, respectively.
How many ways are possible to transport the 9 people using all cars?
7.
a) How many ways can 6 people be lined up to get a bus?
b) If 3 persons insist on following each other, how many ways are possible?
Submit your assignment and have it
checked by your teacher.

Summary
1. The simplest method of counting is done by enumerating all possible outcomes. The tree diagram
helps us list all possible combinations by counting the various paths along the branches provided
the possibilities are at manageable level.
2. The fundamental principle of counting, also known as multiplication rule, states that if the first
activity can be done in n1 ways, a second in n2 ways and so on until the kth
activity in nk ways, the
total ways of doing the activities is n1 x n2 x ... x nk
3. A permutation is an arrangement of all or part of a set of objects.
4. The number of permutation of n distinct objects is n!.
5. The number of permutation of n distinct objects taken r at a time is
nPr =
𝑛!
𝑛−𝑟 !
6. The number of permutation of n distinct objects arranged in a circle is (n -1)!.
7. The number of permutations of n things of which n1 are of one kind , n2 are of a second ...nk of a
kth
kind is
nPn =
𝑛!
𝑛1!𝑛2!…𝑛 𝑘!
8. Combinations are the number of ways of selecting r objects from n without regard to order.
9. The number of combinations of n distinct objects taken r at a time is
nCr =
𝑛!
𝑟! 𝑛−𝑟 !
You have just finished reading the basic concepts and
principles about counting techniques. Let us now
summarize what you learned in module 1.
Job well done!
Congratulations for
completing Module 1

Probability module 1

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Probability module 1