This document summarizes a lecture on logic and problem solving that covered: 1. The basic principle of counting, permutations, and combinations. Permutations involve order while combinations do not. 2. Examples were given to demonstrate the fundamental counting principle and formulas for permutations and combinations. 3. Permutations and combinations with repetition were also discussed along with circular permutations. 4. Exercises were provided to practice applying the concepts through problems involving counting arrangements of letters, books, team selections from groups, and card selections.

1. Lecture week 17
Logic & Problem Solving
Principle of counting

2. Agenda:
 Week 17 lecture coverage
–Basic Principle of counting
–Definition and problems of Permutation
–Definition and problems of Combination

3. Permutations & Combinations

4. Why do we want to study permutation and combination?
• Very often we will find situations where one thing can be done in
different ways.
• In order to find the best way, we need to know how many possible
ways are there in total.
• For example, computer A send message to computer B through the
network. How many possible routes are there? Which one is the best?
A
B

5. • Two students play scissor paper rock
• How many possible ways the play will go?
Scissor Paper Rock
Student 1
Scissor
Paper
Rock
Student 2
Scissor
Paper
Rock
9 Possible ways

6. Fundamental Counting Principle
Fundamental Counting Principle states that if an event has m possible
outcomes and another independent event has n possible outcomes, then there
are m × n possible outcomes for the two events together.

7. Fundamental Counting Principle
Let’s start with a simple example.
A student is to roll a die and flip a coin.
How many possible outcomes will there be?
1H 2H 3H 4H 5H 6H
1T 2T 3T 4T 5T 6T
12 outcomes
𝟔 × 𝟐 = 𝟏𝟐 outcomes

8. Fundamental Counting Principle
Example:
For a college interview, Robert has to choose what to wear from the following:
4 slacks, 3 shirts, 2 shoes and 5 ties. How many possible outfits does he
have to choose from?
𝟒 × 𝟑 × 𝟐 × 𝟓 = 𝟏𝟐𝟎 outfits

9. Permutations
Permutation is an arrangement of items in a particular
order.
ORDER MATTERS!!!
To find the number of Permutations of n items, we can use the
Fundamental Counting Principle or factorial notation (!).

10. The number of ways to arrange the letters USA:
____ ____ ____
Number of choices for first blank? 3
3 2
Number of choices for second blank?
Number of choices for third blank? 3 2 1
3 × 2 × 1 = 6 3! = 3 × 2 × 1 = 𝟔 ways
USA UAS SUA SAU AUS ASU
Permutations

11. To find the number of Permutations of n items chosen r at a time, you
can use the formula
.
0
where n
r
r
n
n
r
p
n 



)!
(
!
60
3
.
4
.
5
2!
5!
)!
3
5
(
!
5
3
5 




p
Permutations

12. A combination lock will open when the right choice of three numbers (from 1
to 30, inclusive) is selected. How many different lock combinations
are possible assuming no number is repeated?
Practice:
Answer Now
Permutations

13. A combination lock will open when the right choice of three numbers (from
1 to 30, inclusive) is selected. How many different lock combinations are
possible assuming no number is repeated?
Practice:
24360
28
.
29
.
30
27!
30!
)!
3
30
(
!
30
3
30 




p
Permutations

14. From a club of 24 members, a President, Vice President, Secretary,
Treasurer and Historian are to be elected. In how many ways can
the offices be filled?
Practice:
Answer Now
Permutations

15. From a club of 24 members, a President, Vice President, Secretary,
Treasurer and Historian are to be elected. In how many ways can
the offices be filled?
Practice:
24 P 5 =
24!
(24−5)!
=
24!
19!
= 24 . 23 . 22 . 21 . 20 = 5,100,480 𝑤𝑎𝑦𝑠
Permutations

16. Combinations
Combination is an arrangement of items in which order
does not matter.
ORDER DOES NOT MATTERS!!!
Since the order does not matter in combinations, there are fewer
combinations than permutations.
The combinations are a "subset "of permutations.

17. To find the number of Combinations of n items chosen r at a
time, you can use the formula
.
0
where
)!
(
!
!
n
r
r
n
r
n
r
C
n




10
2
20
1
.
2
4
.
5
1
.
2
.
1
.
2
.
3
1
.
2
.
3
.
4
.
5
3!.2!
5!
)!
3
5
(
!
3
!
5
3
5







C
Combinations

18. To play a particular card game, each player is dealt five
cards from a standard deck of 52 cards. How many
different hands are possible?
Practice:
Answer Now
Combinations

19. To play a particular card game, each player is dealt five cards
from a standard deck of 52 cards. How many different hands
are possible?
Practice:
960
,
598
,
2
1
.
2
.
3
.
4
.
5
48
.
49
.
50
.
51
.
52
5!47!
52!
)!
5
52
(
!
5
!
52
5
52





C
Combinations

20. A student must answer 3 out of 5 essay questions on a test. In
how many ways can the student select the questions?
Practice:
Answer Now
Combinations

21. A student must answer 3 out of 5 essay questions on a test. In
how many ways can the student select the questions?
Practice:
10
1
.
2
4
.
5
3!2!
5!
)!
3
5
(
!
3
!
5
3
5 




C
Combinations

22. A basketball team consists of two centers, five forwards, and four
guards. In how many ways can the coach select a starting line
up of one center, two forwards, and two guards?
Practice:
Answer Now
Combinations

23. A basketball team consists of two centers, five forwards, and four
guards. In how many ways can the coach select a starting
line up of one center, two forwards, and two guards?
Practice:
Thus, the number of ways to select the starting line up is,
= 𝟐 × 𝟏𝟎 × 𝟔 = 𝟏𝟐𝟎.
2
!
1
!.
1
!
2
1
2 

C
Center:
10
1
.
2
4
.
5
!
3
!.
2
!
5
2
5 


C
Forwards:
6
1
.
2
3
.
4
!
2
!.
2
!
4
2
4 


C
Guards:
2
4
2
5
1
2 .
. C
C
C
Combinations

24. Permutation with Repetition
Permutations: Distinct Objects with Repetition
The number of ordered arrangements of r objects chosen from n
objects, in which the n objects are distinct, and repetition is allowed, is
nr
Let's look at a 3-combination
lock with numbers 0 through
9 where numbers can be
repeated .
There are 10 choices for each of the first
number, second number and third
number.
Number of possible choices = nr
=103 = 1000 choices

25. 1 2
1 2
1 2
The number of permutations of objects
of which are of one kind, are of a
second kind, . . ., and are of a th kind
is given by
!
! ! !
where
k
k
k
n
n n
n k
n
n n n
n n n n
  
   
Permutations Involving n Objects That Are Not
Distinct

26. Permutations Involving n Objects That Are Not Distinct
How many different passwords can be made form the letter of the
following words?
a. Apple =
5!
2!
= 60 passwords
b. Banana =
6!
3! × 2!
= 60 passwords
c. Rearrange = ???
d. Computer = ???
e. Mississippi = ???

27. Circular Permutations Involving n Objects
Circular permutation is the total number of ways in
which 𝒏 distinct objects can be arranged around a fix
circle.
Circular Permutation with 𝒏 objects = ( 𝒏 – 𝟏)!

28. Circular Permutations Involving n Objects
Circular Permutation with 𝒏 objects = ( 𝒏 – 𝟏)!
Example:
In how many different ways can 10 boys
be seated for dinner in a round table ?
No. of ways = (𝟏𝟎 – 𝟏) ! = 𝟗! = 𝟑𝟔𝟐𝟖𝟖𝟎 ways

29. Exercises ….
1 1 x
If then find x.
9! 10! 11!
 
1.
2. Find 𝑥 if 8! × 𝑥! = (𝑥 + 2)! × 6!
3. How many way, 3 chemistry, 2 physics,4 mathematics
book can be arranged such that all books of same
subjects are kept together.
A group consists of 5 girls and 8 boys. In how many
ways can a team of 5 members be selected if the
team has
(i) no girls
(ii) at least three girls.
4.

30. Exercises ….
5.
6.
How many 4-digit arrangements can be created using
the digits 5, 6, 3 and 2 if,
a) the numbers can be repeated
b) the five must be the first number
c) the five must be the first number and the two the
second
In how many ways can 4 cards be selected from a well-shuffled
deck of 52 cards?

31. Summary: Week 17 Lecture
• Principle of Counting
• Permutation
• Combination
• Permutation with repetition
• Circular Permutation

32. What to Expect: Week 17 Tutorials
•Review and practice Permutation and Combination problems through in-
class assignments to acquire them.
•Practice problems to know how concept of counting and selecting can
be useful in solving various mathematical problems.

33. Thank you