Download c1 - School of Computing

Click on the picture
Main Menu
(Click on the topics below)
Combinatorics
Introduction
Equally likely Probability Formula
Counting elements of a list
Counting elements of a sublist
Sum of numbers from 1 to n
Pairs of numbers
Possibility Trees & The Multiplication Rule
Cartesian Product
Subsets of A= {a1, a2,…, an}
3 digit numbers with distinct digits
Relations from A to B
3 digit +ve odd integers with distinct digits
Combinatorics
Counting the number of possible
outcomes.
Counting the number of ways a
task can be done.
Sanjay Jain, Lecturer,
School of Computing
Click on the picture
Introduction


Multiplication Rule
Examples
Sanjay Jain, Lecturer,
School of Computing
Combinatorics
Counting
Probability
 To say that a process is random means that when it
takes place, one out of a possible set of outcomes
will occur. However it is in general impossible to
predict with certainty which of the possible
outcomes will occur.
 A sample space is the set of all possible outcomes
of a random experiment.
 An event is a subset of a sample space.
Example
Tossing two coins.
Sample Space: {HH, HT, TH, TT}
Event: At least one head: {HH, HT, TH}
END OF SEGMENT
Equally Likely Probability Formula
Suppose S is a sample space in which all outcomes
are equally likely.
Suppose E is an event in S.
Then the probability of E, denoted by Pr(E) is
Pr( E ) 
# (E)
# (S )
Notation: For a set A, #(A) denotes the number of
elements in A.
Sometimes n(A) or || A || is also used for #(A).
Sometimes Prob(E) or P(E) is also used for Pr(E).
Example
Consider the process of drawing a card from a pack of
cards.
What is the probability of drawing an Ace?
Assume drawing any card is equally likely.
Sample Space: S={SA, S2, S3, …., HA, H2,….}.
Event: E={SA, HA, DA, CA}
#(S)=52
#(E)=4
Pr(E)=4/52
END OF SEGMENT
Counting The Elements of A List
How many integers are there from 8 through 15?
8
9
10
11
12
1
2
3
4
5
13 14 15
6
7
8
Theorem
If m and n are integers and m  n then there are
n-m+1 integers from m to n (both inclusive).
Proof:
m
m+1 m+2 ………………… n
m+0 m+1 m+2 …………………m+(n-m)
1
2
3
……………… (n-m)+1
END OF SEGMENT
Counting Elements of a Sublist
How many 3 digit positive integers are divisible by 5?
100
105 ………. 995
20*5 21*5 ……… 199*5
20
21 ……….. 199
199 -20+1=180
END OF SEGMENT
Floors and Ceilings
w denotes the largest integer w.
For example: 6.9 = 6; -9.2 = -10;
w denotes the smallest integer  w.
For example: 6.9 =7;
-9.2 = -9;
9 = 9
9 = 9
END OF SEGMENT
Sum of numbers from 1 to n
Theorem: 1+2+….+n = n(n+1)/2
Proof
We show this by induction on n.
For n=1, the above is clearly true.
Suppose the theorem holds for n = k.
We show the theorem for n = k+1.
1 + 2 + … + k + (k+1)
= [k (k + 1) / 2] + (k+1)
= (k + 2) ( k + 1) / 2
= (k + 1) (k + 2) / 2
END OF SEGMENT
Pairs of numbers:
How many distinct pairs of numbers (i,j) satisfy the
property 1  i < j  n?
For any i, 1  i < n, the number of j’s which satisfy
1  i < j  n, is n - i.
Thus, the number of distinct pairs of numbers (i,j) that
satisfy the property 1  i < j  n is
n 1
ni
= ( n  1) * n 
i 1
(n  1)n
(n  1) * n 
2
n 1
i
i 1
=
( n  1) n
2
=
END OF SEGMENT
Possibility Trees
Coin Toss:
Toss
H
T
2 ways
H
2 x 2 ways
T
H
T
END OF SEGMENT
The Multiplication Rule
Theorem: If an operation (or job) consists of k tasks
(or steps), T1, T2,…, Tk, performed one after
another and
T1
can be done in n1 ways
T2 can be done in n2 ways (irrespective of how
T1 is done)
….
Tk can be done in nk ways (irrespective of how
T1 ... Tk-1 are done)
Then, the entire operation can be done in n1* n2* ….*
nk ways.
The Multiplication Rule
Theorem: If an operation (or job) consists of k tasks
(or steps), T1, T2,…, Tk, performed one after
another and
Ti can be done in ni ways (irrespective of how
T1 ... Ti-1 are done)
Caution: Note the independence assumption. One
cannot use the multiplication rule unless the
independence assumption holds.
END OF SEGMENT
Cartesian Product
How many elements are there in A x B?
A= {a1, a2,…., an}
B= {b1, b2,…., bm}
Recall: A X B = {(a,b) : a A and b  B}.
Cartesian Product
Job: select an element of A X B.
 T1: Select an element a of A
 T2: Select an element b of B
 (this gives us an element (a,b) of A X B)
 T1 can be done in n ways
 T2 can be done in m ways (irrespective of how T1
is done)
 by the multiplication rule, the job can be done in
n*m ways.
The number of elements of A x B is n*m
END OF SEGMENT
Subsets of A= {a1, a2,…, an}
How many subsets of A={a1, a2,…, an} are there?
Job: select a subset of A.
 T1: either select or not select a1
 T2: either select or not select a2
 ….

Tn: either select or not select an
Each of these tasks can be done in two ways
(irrespective of how the earlier tasks are done).
Thus the number of ways of doing the job is 2n.
Therefore, the number of subsets of A is 2n.
END OF SEGMENT
3 digit numbers with distinct digits
How many 3 digit numbers with distinct digits are there?
3 digit numbers with distinct digits
How many 3 digit numbers with distinct digits are there?
 T1: Select the hundred’s digit
 T2: Select the ten’s digit
 T3: Select the unit’s digit
3 digit numbers with distinct digits
How many 3 digit numbers with distinct digits are there?

T1: Select the hundred’s digit

T1 can be done in 9 ways (digit 0 cannot be
selected)
3 digit numbers with distinct digits
How many 3 digit numbers with distinct digits are there?

T2: Select the ten’s digit

T2 can be done in 9 ways (irrespective of how T1
was done). You cannot select the digit chosen in
T1
3 digit numbers with distinct digits
How many 3 digit numbers with distinct digits are there?

T3: Select the unit’s digit

T3 can be done in 8 ways (irrespective of how
earlier tasks were done). You cannot select the
digit chosen in T1 and T2
3 digit numbers with distinct digits
How many 3 digit numbers with distinct digits are there?
 T1 can be done in 9 ways
 T2 can be done in 9 ways
 T3 can be done in 8 ways
Therefore, the total number of 3 digit numbers with distinct
digits are 9*9*8
END OF SEGMENT
Relations From A to B
How many different relations are there from A to B?
A={a1, a2,…., an}, B={b1, b2,…., bm}
T(i,j) : select or not select (ai,aj) as a member of R.
(1  i n and 1  j m)
Note that the total number of tasks is n*m.
Each T(i,j) can be done in 2 ways.
Thus all the tasks can be done in 2n*m ways
Total number of relations is: 2n*m
Relations From A to B
How many different relations are there from A to B?
A={a1, a2,…., an}, B={b1, b2,…., bm}
Another Method:
A relation is a subset of A X B.
Number of elements in A X B = n*m
number of subsets of A X B = 2n*m
END OF SEGMENT
Be careful in using the Multiplication Rule
How many 3 digit +ve odd integers have distinct digits?
 T1: Select the hundred’s digit
 T2: Select the ten’s digit
 T3: Select the unit’s digit
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
 T1: Select the hundred’s digit

T1 can be done in 9 ways
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
 T2: Select the ten’s digit

T2 can be done in 9 ways (irrespective of how T1
was done). You cannot select the digit chosen in
T1
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
 T3: Select the unit’s digit

T3 can be done in ? ways (the number of ways is
either 3 or 4 or 5 depending on how exactly T1
and T2 were done).
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
 T1 can be done in 9 ways
 T2 can be done in 9 ways
 T3 can be done in ? ways
Therefore, the Multiplication Rule may not always be
applicable.
However, for this problem one can use the Multiplication
Rule by reordering tasks.
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
(reordering tasks)
 T1: Select the unit’s digit
 T2: Select the hundred’s digit
 T3: Select the ten’s digit
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
(reordering tasks)
 T1: Select the unit’s digit

T1 can be done in 5 ways
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
(reordering tasks)
 T2: Select the hundred’s digit

T2 can be done in 8 ways (irrespective of how T1
was done).
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
(reordering tasks)
 T3: Select the ten’s digit

T3 can be done in 8 ways (irrespective of how T1
and T2 are done).
3 digit +ve odd integers
How many 3 digit +ve odd integers have distinct digits?
(reordering tasks)
 T1 can be done in 5 ways
 T2 can be done in 8 ways
 T3 can be done in 8 ways
Therefore, the total number of 3 digit +ve odd integers
with distinct digits is 5*8*8
END OF SEGMENT
Symmetric Relations
Suppose A ={a1,a2,…,an}.
How many symmetric relations can be defined on A?
We will show that it is 2n(n+1)/2
Recall: for a relation to be symmetric, for each i, j, either
both (ai,aj) and (aj,ai) are in R or both are not in R.
Divide the job of selecting a symmetric relation R into the
following tasks.
Si (for 1  i  n)
Either select or not select (ai,ai) in R
T(i,j) (for 1  i < j  n)
Either select or not select both (ai,aj) and (aj,ai) in R
Note that the number of different T(i,j) 's are (n-1)n/2
Symmetric Relations
Si (for 1i n)
Either select or not select (ai,ai) in R
T(i,j) (for 1i<j n)
Either select or not select both (ai,aj) and (aj,ai)
in R
Each Si and T(i,j) can be done in exactly 2 ways.
Thus the total number of symmetric relations on A are
(2*2*…*2) * (2*2*….*2)
(there are n 2’s in the first group, and ((n-1)n/2)
2’s in the second group)
=2n*2n(n-1)/2
=2n(n+1)/2
END OF SEGMENT
Simple Graphs
How many simple undirected graphs are there with n
vertices?
This is similar to symmetric relations except that Si’s are not
there.
T(i,j) (for 1  i < j  n)
Either select or not select the edge {vi,vj} (= {vj,vi})
Note that the number of different T(i,j) 's are (n-1)n/2
Each T(i,j) can be done in exactly 2 ways.
Thus the total number of simple graphs is
2*2*….*2
(there are ((n-1)n/2) 2’s )
=2n(n-1)/2
END OF SEGMENT
Click on the picture
Summary


Multiplication Rule
Remember the conditions under which multiplication
rule is applicable, specially note the independence
assumption
Click on the picture
Follow-Up



Explain assignments.
List books, articles, electronic sources.
If appropriate, give an introduction to the next lecture in
the series.