DM unit 3

1. Unit VI
Discrete Structures
Permutations and Combinations
SE (Comp.Engg.)

2. • BOTH
• PERMUTATIONS AND COMBINATIONS
• USE A COUNTING METHOD CALLED
FACTORIAL

3. • Lets start with a simple example.
• A student is to roll a die and flip a
coin. How many possible outcomes will
there be?

4. • 1H 2H 3H 4H 5H 6H
• 1T 2T 3T 4T 5T 6T

5. • The number of ways to arrange the letters
ABC:
____ ____ ____
Number of choices for first blank? 3 ____ ____
3 2 ___
Number of choices for second blank?
Number of choices for third blank? 3 2 1

6. Permutations
• A Permutation is an arrangement of
items in a particular order.
The number of Permutations of n items
chosen r at a time, is given by the
formula
.
0
where n
r
r
n
n
r
p
n 



)!
(
!

7. • Arrange 2 alphabets from a,b,c.
• ab, ba, ac,ca,bc,cb
• 3P2=3!/(3-2)!=6

8. • 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?

9. 24360
28
*
29
*
30
)!
3
30
(
!
30
3
30 




27!
30!
p

10. CIRCULAR
PERMUTATIONS
When items are in a circular
format, to find the number of
different arrangements,
divide:
n! / n

11. • Six students are sitting around a circular table
in the cafeteria. How many different seating
arrangements are there?

12. •6!  6 = 120

13. Combination
• A Combination is an arrangement of
items in which order does not matter.
.
0
where n
r
r
n
r
n
r
C
n




)!
(
!
!
The number of Combinations of n
items chosen r at a time, is given by
the formula

14. • 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?

15. 960
,
598
,
2
1
*
2
*
3
*
4
*
5
48
*
49
*
50
*
51
*
52
)!
5
52
(
!
5
!
52
5
52





5!47!
52!
C

16. • A student must answer 3 out of 5 essay
questions on a test. In how many
different ways can the student select
the questions?

17. 10
1
*
2
4
*
5
)!
3
5
(
!
3
!
5
3
5 




3!2!
5!
C

18. Product and Sum Rules
• Product Rule:
• If we need to perform procedure 1 AND
procedure 2.
There are n1 ways to perform procedure 1 and n2
ways to perform procedure 2.
• There are n1•n2 ways to perform procedure 1 AND
procedure 2.

19. • Sum Rule:
If need to perform either procedure 1 OR
procedure 2. There are n1 ways to perform
procedure 1 and n2 ways to perform procedure 2.
• There are n1+n2 ways to perform procedure 1 OR
procedure 2.
• This “OR” is an “exclusive OR.” One choice or the other, but
not both.

20. • How many vehicle number plates can be
made if each plate contains two different
letters followed by three different digits

21. Two different letters are made in 26P2 ways.
3 different digits are combined in 10P3ways
Total no. of number plates= 26P2 * 10P3
=26!*10!/(24!*7!)
=26*25*10*9*8
=468000

22. • An 8 member team is to be formed from a
group of 10 men and 15 women. In how many
ways can the team be chosen if :
(i) The team must contain 4 men and 4 women
(ii) There must be more men than women
(iii) There must be at least two men

23. The team must contain 4 men and 4 women =
286650
(ii) There must be more men than women =
(iii) There must be at least two men

24. • Passwords consist of character strings of 6 to 8
characters. Each character is an upper case
letter or a digit. Each password must contain
at least one digit.
• How many passwords are possible?

25. • Total number is
• # passwords with 6 char. + # passwords
with 7 char. + # pws 8 char.
• (=P6+P7+P8).

26. • P6: # possibilities without constraint : 366
• # exclusions is # passwords without any
digits is 266
• And so, P6 = 366-266

27. • Similarly, P7 = 367-267 and P8 = 368-268

28. • P = P6+P7+P8 = 366-266 + 367-267 + 368-2

29. • How many bit-strings of length 8 either begin
with 1 or end with 00?

30. • A = 8-bit strings starting with 1
• |A| = # of 8-bit strings starting with 1 is 27
• B = 8-bit strings starting with 00
• |B| = # of 8-bit strings ending with 00 is 26
• # of bit-strings begin with 1 and end with 00 is
25.

31. • # of 8-bit strings starting with 1 or ending with
00 is
• 27+ 26- 25

32. • |AB| = |A| + |B| - |AB|
• Inclusion–Exclusion Principle

33. 33
Permutations with
non-distinguishable objects
• The number of different permutations of n
objects, where there are non-distinguishable
objects of type 1, non-distinguishable objects
of type 2, …, and non-distinguishable objects
of type k, is
i.e., C(n, )C(n- , )…C(n- - -…- , )
1 2
!
! !... !
k
n
n n n
1
n
2
n
k
n
1
n 1
n 2
n 1
n 2
n 1
k
n  k
n
1 2 ... k
n n n n
   

34. • How many different strings can be made by
reordering the letters of the word
OFF

35. • 3!/2!=3
• OFF
• FFO
• FOF

36. • ONE
• OEN
• NEO
• NOE
• ENO
• EON

37. • How many different strings can be made by
reordering the letters of the word
SUCCESS

38. Generating Permutations
Lexicographic method

39. • For the following 4
• combinations from the set f= {1;2;3;4;5;6;7}
find the combination that immediately follows
them in lexicographic order
1234 is followed by
3467 is followed by
4567 is followed by

40. What is probability?
• Probability is the measure of how likely an
event or outcome is.
• Different events have different probabilities!

41. How do we describe probability?
• You can describe the probability of an event with the
following terms:
– certain (the event is definitely going to happen)
– likely (the event will probably happen, but not definitely)
– unlikely (the event will probably not happen, but it might)
– impossible (the event is definitely not going to happen)

42. • probabilities are expressed as fractions.
– The numerator is the number of ways the event
can occur.
– The denominator is the number of possible events
that could occur.

43. 6.43
Random Experiment…
• …a random experiment is an action or process
that leads to one of several possible
outcomes. For example:
Experiment Outcomes
Flip a coin Heads, Tails
Exam Marks Numbers: 0, 1, 2, ..., 100
Assembly Time t > 0 seconds
Course Grades F, D, C, B, A, A+

44. What is the probability that I will choose a
red marble?
• In this bag of marbles, there are:
– 3 red marbles
– 2 white marbles
– 1 purple marble
– 4 green marbles

45. Probability example
• Sample space: the set of all possible outcomes.
• Probabilities: the likelihood of each of the possible outcomes
(always 0 P 1.0).
)
(
1
)
(
)
(
1
)
(
1
)
(
)
(
1
)
(
0
E
P
E
P
E
P
E
P
E
P
E
P
E
P









46. 6.46
Probabilities…
• List the outcomes of a random experiment…
• This list must be exhaustive, i.e. ALL possible
outcomes included.
• Die roll {1,2,3,4,5} Die roll {1,2,3,4,5,6}
• The list must be mutually exclusive, i.e. no
two outcomes can occur at the same time:
• Die roll {odd number or even number}
• Die roll{ number less than 4 or even number}

47. • A and B are independent if and only if
P(A&B)=P(A)*P(B)
• A and B are mutually exclusive events:
P(A or B) = P(A) + P(B)

48. 6.48
Events & Probabilities…
• The probability of an event is the sum of the
probabilities of the simple events that constitute
the event.
• E.g. (assuming a fair die) S = {1, 2, 3, 4, 5, 6} and
• P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6
• Then:
• P(EVEN) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 =
3/6 = 1/2

49. Female
Low
Male
e.g. 1. The following table gives data on the type of car, grouped by
petrol consumption, owned by 100 people.
4
21
23
7
33
12
High
Medium
One person is selected at random.
L is the event “the person owns a low rated car”
F is the event “a female is chosen”.
Find (i) P(L) (ii) P(F  L) (iii) P(F| L)
100
Total

50. (i) P(L) =
Solution:
Find (i) P(L) (ii) P(F  L) (iii) P(F L)
100
4
21
23
Female
7
33
12
Male
High
Medium
Low
20 20
7

7
35
100
(ii) P(F  L) =
23
Total
100
(iii) P(F L) 
23
Notice that
P(L)  P(F L)
35
23
20
7


100
23

So, P(F  L) = P(F|L)  P(L)
= P(F  L)
35
5
1

51. 45
R
F
e.g. 2. I have 2 packets of seeds. One contains 20 seeds and
although they look the same, 8 will give red flowers and 12 blue.
The 2nd packet has 25 seeds of which 15 will be red and 10 blue.
Draw a Venn diagram and use it to illustrate the conditional
probability formula.
Solution:
15
10
12
P(R  F) =
P(F) =
P(R F) =
8
8
45
20
8
45
20
20
8

P(R  F) = P(R|F)  P(F)
So,
P(R F)  P(F) =

20
45
45
8

1
1
Let R be the event “ Red flower ” and
F be the event “ First packet ”

52. Bayes’ Rule: derivation
)
(
)
&
(
)
/
(
B
P
B
A
P
B
A
P 
• Definition:
Let A and B be two events with P(B)  0. The
conditional probability of A given B is:

53.      
       
B
A
P
B
P
B
A
P
B
P
B
A
P
B
P
A
B
P
|~
~
|
|
|



54. Example
• Three jars contain colored balls as described in the table
below.
– One jar is chosen at random and a ball is selected. If the ball is red,
what is the probability that it came from the 2nd jar?
Jar # Red White Blue
1 3 4 1
2 1 2 3
3 4 3 2

55. Example
• We will define the following events:
– J1 is the event that first jar is chosen
– J2 is the event that second jar is chosen
– J3 is the event that third jar is chosen
– R is the event that a red ball is selected

56. Example
• The events J1 , J2 , and J3 mutually exclusive
– Why?
• You can’t chose two different jars at the same time
• Because of this, our sample space has been
divided or partitioned along these three
events

57. Venn Diagram
• Let’s look at the Venn Diagram

58. Venn Diagram
• All of the red balls are in the first, second, and
third jar so their set overlaps all three sets of
our partition

59. Finding Probabilities
• What are the probabilities for each of the
events in our sample space?
• How do we find them?
     
B
P
B
A
P
B
A
P |



60. Computing Probabilities
• Similar calculations
show:
     
8
1
3
1
8
3
| 1
1
1 



 J
P
J
R
P
R
J
P
     
     
27
4
3
1
9
4
|
18
1
3
1
6
1
|
3
3
3
2
2
2










J
P
J
R
P
R
J
P
J
P
J
R
P
R
J
P

61. Venn Diagram
• Updating our Venn Diagram with these
probabilities:

62. Where are we going with this?
• Our original problem was:
– One jar is chosen at random and a ball is selected.
If the ball is red, what is the probability that it
came from the 2nd jar?
• In terms of the events we’ve defined we want:
   
 
R
P
R
J
P
R
J
P

 2
2 |

63. Finding our Probability
   
 
 
     
R
J
P
R
J
P
R
J
P
R
J
P
R
P
R
J
P
R
J
P









3
2
1
2
2
2 |

64.    
     
17
.
0
71
12
27
4
18
1
8
1
18
1
|
3
2
1
2
2




































R
J
P
R
J
P
R
J
P
R
J
P
R
J
P