The document discusses permutations and combinations. It provides examples of calculating permutations and combinations for different scenarios like selecting committees from a group of people and arranging books on a shelf. Formulas for permutations (nPr) and combinations (nCr) are given. Order matters for permutations but not for combinations. The key difference between the two is explained.

1. BY UNSA SHAKIR

2. The following data give the total number of iPods sold
by a mail order company on each of 30 days.
Construct a frequency table.
Find the mean, variance and standard deviation, mode
and median.
23 14 19 23 20 16 27 9 21 14
22 13 26 16 18 12 9 26 20 16
8 25 11 15 28 22 10 5 17 21

3. FCP (fundamental counting priciple)
Factorial Notation
Permutation
Combination

4. Fundamental Counting Principle can be used
determine the number of possible outcomes
when there are two or more characteristics .

5. PROBLEM
 Liza brought 3 different pairs of pants and 4 shirts
in a camp. How many combinations of a shirt and a
pair of pants can she choose from to wear?
• Lets name the shirts of Liza as Shirts A, B, C, and D.
Lets also name her pants as Pants 1, 2, and 3.
• Let us now enumerate the possible combinations
that Liza can wear

6. The answer is there are 12 possible
combinations.
We found the answer by drawing a diagram.
But what if there are large numbers given in the
problem (say, 179 pairs of pants and 83 shirts)?

7. If an independent event can occur in m ways,
another independent event can occur in n
ways, and another in p ways, then the total
number of ways that all events can occur
simultaneously is
n(E) = m∙n∙p ways

8. In the given problem,
n(E) = 4 x 3 = 12 ways

9. 9
Example:
When buying a PC system, you have the choice of
 3 models of the basic unit: B1, B2, B3 ;
 2 models of keyboard: K1, K2 ;
 2 models of printer: P1, P2 .
Question:
How many distinct systems can be purchased?
Solution: The number of distinct systems is: 3∙2∙2=12

10. Fundamental Counting Principle
Lets 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

11. Fundamental Counting Principle
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

12.  A PIN is a sequence of any 4 digits (repetitions
allowed); e.g., 5279, 4346, 0270.
Question. How many different PINs are possible?
Solution. Choosing a PIN is a 4-step operation:
 Step 1: Choose the 1st symbol (10 different ways).
 Step 2: Choose the 2nd symbol (10 different ways).
 Step 3: Choose the 3rd symbol (10 different ways).
 Step 4: Choose the 4th symbol (10 different ways).
Based on the multiplication rule,
10∙10∙10∙10 = 10,000 PINs are possible.
EXAMPLE

13. 13
Example
Consider the problem of choosing PINs
but now repetitions are not allowed.
Question. How many different PINs are possible?
Solution. Choosing a PIN is a 4-step operation:
 Step 1: Choose the 1st symbol (10 different ways).
 Step 2: Choose the 2nd symbol (9 different ways).
 Step 3: Choose the 3rd symbol (8 different ways).
 Step 4: Choose the 4th symbol (7 different ways).
Based on the multiplication rule,
10∙9∙8∙7 = 5,040 PINs are possible.

14. • In a certain class, there are 23 majors in Psychology, 16 majors in
English and 7 students who are majoring in both Psychology and
English.
• A) If there are 50 students in the class, how many students are
majoring in neither of these subjects?
• B) How many students are majoring in Psychology alone?
example
Solve Using Venn-Diagram

15. Solution:
We introduce the following principle
of counting that can be illustrated using
a Venn-Diagram.
N( A U B) = n(A) + n(B) – n(A B)
This statement says that the number of
elements in the union of two sets A and
B is the number of elements of A
added to the number of elements of B
minus the number of elements that are
in both A and B.

A B

16. Both Psych
and English
7 students in this region
23 students here
• N(P U E) =
n(P)+n(E)-n(P E)
• 23 + 16 – 7 = 32
a.
b. 23 – 7 = 16

17. A marketing survey of a group of kids indicated that 25 owned a
Nintendo DS and 15 owned a Wii. If 10 kids had both a DS and a
Wii, how many kids interviewed have a DS or a Wii?
Let D represent the set of kids with a DS, and let W represent
the set of kids with a Wii.
Solution:
Number of kids with a DS or a Wii: 30.
Example

18. DS Wii
Both DS and
Wii
10
25
• n(D ∪ W) = n(D) + n(W)
− n(D ∩ W)
• n(D ∪ W) = 25 + 15 − 10
= 30

19. More problems…
• How many different ways can a team consisting of 28 players
select a captain and an assistant captain?
• Solution: Operation 1: select the captain. If all team members
are eligible to be a captain, there are 28 ways this can be done.
• Operation 2. Select the assistant captain. Assuming that a player
cannot be both a captain and assistant captain, there are 27 ways
this can be done, since there are 27 team members left who are
eligible to be the assistant captain.
• Then, using the multiplication principle there are (28)(27) ways
to select both a captain and an assistant captain. This number
turns out to be 756.

20. The factorial of an integer k is the product of
all integers from 1 to k.
This is usually denoted as k! read as “k
factorial”

21. Examples:
 7! = 7∙6∙5∙4∙3∙2∙1 = 5040
 5! = 5∙4∙3∙2∙1 = 120

22. The factorial operation is NOT distributive.
 Ex. (5-3)! ≠ 5! – 3!
The factorial operation cannot be performed
on non-integer numbers.
 Ex. (2.5)!, (√6)!

23. The factorial notation precedes multiplication
and/or division.
 Ex. 5!3! ≠ (5∙3)!
 8! / 2! ≠ (8/2)!
0! = 1

24. n! = n(n-1)!
= n(n-1)(n-2)!
= n(n-1)(n-2)(n-3)…(n-k+1)(n-k)!
This property is very crucial in making
cancellations in factorial expressions.

25. Example
= 4896

26. BOTH
PERMUTATIONS AND
COMBINATIONS
USE A COUNTING METHOD
CALLED FACTORIAL.

27. A permutation is an ordered arrangement
of objects.
It tells us how many possible orders there can
be given a number of objects.
In permutation, if all objects are distinct,
then they cannot be repeated.

28. Permutation is usually denoted by
nPr
read as “the permutation of n objects taken r at
a time”

29. The general formula for permutation is
nPr = n(n-1)(n-2)…(n-r+1)
or

30. n is the total number of objects
r is the number of objects in consideration
In the problem given, r=n
If r=n, the equation becomes

31. Example: Permutations
Evaluate each permutation.
a) 5P3 b) 6P6
5 3
5! 5!
a) 60
(5 3)! 2!
P   

6 6
6! 6!
b) 720
(6 6)! 0!
P   

Solution

32. Permutations
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:
2436028*29*30
)!330(
!30
330 


27!
30!
p

33. Permutations
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:
480,100,520*21*22*23*24
)!524(
!24
524




19!
24!
p

34. 1. How many ways can you choose your top 2 senators
from a list of 35 candidates?
2. we only want the different arrangements for groups of 3
out of 8?
3. Five books are chosen from a group of ten, and put on a
bookshelf. How many possible arrangements are there?
4. Choose 4 books from a group of 7 and arrange them on
a shelf. How many different arrangements are there?
Example

35. Example: Building Numbers From a Set of Digits
How many four-digit numbers can be written using
the numbers from the set {1, 3, 5, 7, 9} if repetitions
are not allowed?
5 4
5! 5!
120.
(5 4)! 1!
P   

Solution
Repetitions are not allowed and order is important, so
we use permutations:

36. Example
 How many ways can three books be arranged in
a shelf?
Possible Arrangements
ABC
ACB
BAC
BCA
CAB
CBA

37. Take Note:
 Since order is essential, ABC is different from
ACB, and all other similar instances.
• Six arrangements or 3! = 3x2x1 = 6

38. How many ways can 4 books be arranged on a shelf?
4! or 4x3x2x1 or 24 arrangements
Here are the 24 different arrangements:
ABCD ABDC ACBD ACDB ADBC ADCB
BACD BADC BCAD BCDA BDAC BDCA
CABD CADB CBAD CBDA CDAB CDBA
DABC DACB DBAC DBCA DCAB DCBA

39. If we were asked how many permutations we
can get if we rearrange the letters of the word
“cat”, we get 3!
What if we rearrange the letters of the word
“dad”?

40. Permutations of the word “dad”
 dad
 dda
 add
There are 3! permutations for the word “dad”.
This leaves us with 3 distinguishable
permutations.

41. Example: IDs
How many ways can you select two letters followed
by three digits for an ID if repeats are not allowed?
Solution
26 2 10 3 650 720 468,000P P   
There are two parts:
1. Determine the set of two letters.
2. Determine the set of three digits.
Part 1 Part 2

42. How many ways can a student government select a president,
vice president, secretary, and treasurer from a group of 6
people?
This is the equivalent of selecting and arranging 4 items
from 6.
= 6 • 5 • 4 • 3 = 360
Divide out common factors.
There are 360 ways to select the 4 people.
Substitute 6 for n and 4 for r in

43. • How many ways can a student government select a
president, vice president, secretary, and treasurer from
a group of 6 people?
• How many ways can a stylist arrange 5 of 8 vases from
left to right in a store display?
• How many ways can a 2-digit number be formed by
using only the digits 5–9 and by each digit being used
only once?

44. 1. Six different books will be displayed in the library
window. How many different arrangements are there?
2. The code for a lock consists of 5 digits. The last number
cannot be 0 or 1. How many different codes are
possible? 80,000
720
3. The three best essays in a contest will receive gold, silver,
and bronze stars. There are 10 essays. In how many ways
can the prizes be awarded?
4. In a talent show, the top 3 performers of 15 will advance
to the next round. In how many ways can this be done?
455
720

45. A combination is an arrangement of objects with
no respect to order. Combinations are basically
“selections”.
Example.
 Find the possible permutations and combinations
of the word “cat”

46. For permutations:
CAT
ACT
TAC
CTA
ATC
TCA
There are 6 permutations for the word “cat”.

47. For combinations,
Since order is regardless, then
CAT = CTA= ACT = ATC = TAC =TCA
which gives us only 1 combination.

48. What if we take only two letters from the word
“cat”?
How many permutations and combinations
are there?

49. For permutations,
 According to the formula, there are 3P2 = 6
permutations, which are
AC
CA
TA
AT
CT
TC

50. For combinations,
Again, order is not essential, soAC=CA,
TC=CT, andAT=TA.
This gives us only 3 combinations.

51. The general formula for a combination is
and it is read “the combination of n objects
taken r at a time”

52. Example: Combinations
Evaluate each combination.
a) 5C3 b) 6C6
5 3
5! 5!
a) 10
3!(5 3)! 3!2!
C   

6 6
6! 6!
b) 1
6!(6 6)! 6!0!
C   

Solution

53. Combinations
A student must answer 3 out of 5
essay questions on a test. In how many
different ways can the student select
the questions?
Practice:
10
1*2
4*5
)!35(!3
!5
35 


3!2!
5!
C

54. Example: Finding the Number of Subsets
Find the number of different subsets of size 3
in the set {m, a, t, h, r, o, c, k, s}.
Solution
A subset of size 3 must have 3 distinct elements,
so repetitions are not allowed. Order is not
important.
9 3
9! 9!
84
3!(9 3)! 3!6!
C   


55. Example: Finding the Number of Poker Hands
A common form of poker involves hands (sets) of five
cards each, dealt from a deck consisting of 52 different
cards. How many different 5-card hands are possible?
Solution
Repetitions are not allowed and order is not
important.
52 5
52! 52!
2,598,960
5!(52 5)! 5!47!
C   


56. 1. Find the number of ways of getting just a
pair when randomly getting three cards from a
standard deck.
2. A committee of 3 students must be selected
from a group of 5 people. How many possible
different committees could be formed?
Example:

57. 1. Find the number of ways of getting at least a
pair when randomly getting three cards from a
standard deck.
2.How many possible committees of 2 people
can be selected from a group of 8?
Example:

58. 1. In a party, there are 53 guests who shook
hands with one another exactly once. How many
handshakes took place?
2. How many committees of 4 students could be
formed from a group of 12 people?
Example:

59. • An amusement park has 27 different rides. If you
have 21 ride tickets, how many different combinations
of rides can you take? 296,010
Pop’s Pizza offers 4 types of meat and 3 types of
cheese. In how many ways could a pizza with two
meats, different or double of the same meat, and one
cheese be ordered? 30

60. Combinations
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:
2
!1!1
!2
12 C
Center:
10
1*2
4*5
!3!2
!5
25 C
Forwards:
6
1*2
3*4
!2!2
!4
24 C
Guards:
Thus, the number of ways to select the
starting line up is 2*10*6 = 120.
22512 * CCC 4*

61. Guidelines on Which Method to Use
Permutations Combinations
Number of ways of selecting r items out of n items
Repetitions are not allowed
Order is important. Order is not important.
Arrangements of n items
taken r at a time
Subsets of n items taken r
at a time
nPr = n!/(n – r)! nCr = n!/[ r!(n – r)!]
Clue words: arrangement,
schedule, order
Clue words: group,
sample, selection