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CSC224/226 -Packet 6- Counting & Graph Theory
Packet #6: Counting & Graph Theory
Applied Discrete Mathematics
Table of Contents
Counting
Graph Theory
Exam Study Sheet
Pages 1-8
Pages 9-16
Page 17
CSC224/226 -Packet 6- Counting & Graph Theory
Counting Information
I.
Product Rule:
ABC = A*B*C
II.
Union Rule:
AB = A + B - AB
III.
Inclusion/Exclusion Principle:
The Union Rule applied inductively to: A 1A 2 …A n 
IV.
Selection with Replacement
r
A.
Permutations:
n
B.
Combinations:
C(n+r-1,r) or C(n+r-1,n-1)
V.
Selection without Replacement
A.
Permutations:
P(n,r)
B.
Combinations:
C(n,r)
VI.
Pigeonhole Principle
If (k+1) or more objects are placed in k boxes, then there is at least one box
containing two or more objects.
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CSC224/226 -Packet 6- Counting & Graph Theory
Counting
Product Rule
ABC = A*B*C
Examples
Example 1:
A = {1,2,3} B = {a,b} C = {a,b,g,d}
How many 3-character words with 1st character from A, 2nd from B, 3rd from C?
Answer: A*B*C = 3*2*4 = 24
Example 2:
3 Independents, 2 Democrats, 4 Republicans
How many ways to form 3-person committee with one member from each
party?
Method: Count number of triples <Independent, Democrat, Republican>
Answer: 3*2*4 = 24
Union Rule
AB = A + B - AB
Example
S = {a,b,c}
*
How many 3-letter words in S start with a or end in b?
Method:
Let A = 3 letter words starting with "a"
Let B = 3 letter words ending with "b"
Find: AB
Answer:
A = (1) (3) (3) = 1*3*3 = 9
B = (3) (3) (1) = 3*3*1 = 9
AB = (1) (3) (1) = 1*3*1 = 3
AB = 9 + 9 - 3 = 15
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Inclusion/Exclusion Principle
To calculate A 1A 2 …A n ,
1. Calculate size of all possible intersections
2. Add results from  of odd numbered sets.
3. Subtract results from  of even numbered sets.
Ie. A 1A 2 A 3  = A 1 + A 2  + A 3  - A 1A 2  - A 1A 3  - A 2 A 3  +
A 1 A 2  A 3 
Example:
How many integers in {1,…,1000} are divisible by either 2, 3, or 5?
A = divisible by 2
B = divisible by 3
C = divisible by 5
Find: ABC
A = {2,4,…,1000} = 500
B = {3,6,…,999} = 333
C = {5,10,…,1000} = 200
AB = {6,12,18,…,996} = 166
AC = {10,20,…,1000} = 100
BC= {15,30,…,990} = 66
ABC = {30,60,…,990} = 33
So, ABC = 500 + 333 + 200 - 166 - 100 - 66 + 33 = 734
Selection with Replacement
Permutations (Order Matters), with replacement: n
r
Choose r times from a set of n items with replacement and order matters.
Drawing with replacement (sequence can repeat items):
If there are n distinct objects and we draw r times. Each drawing
has n distinct possibilities. If order matters, then by the product
r
rule we have n x n x ... x n possibilities = n .
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Examples
Example 1: Create a sequence of 10 numbers by rolling a die 10 times. How many
possible sequences?
Method: Choose 10 times from a set of 6.
10
Answer: 6
Example 2: S = {0,1,…,7}
How many strings of length 5 are there in S * ?
Method: Choose 5 times from a set of 8.
5
Answer: 8
Combinations (Order Does Not Matter), with replacement: C(n+r-1,r)
Choose r times from a set of n items with replacement and order does not matter.
The number of possible outcomes is written as C(n+r-1,r).
Combinations are discussed in more detail later.
Example:
You have 4 different kinds of cookies. How many different ways can 6 cookies be
chosen?
Method: Choose 6 times from a set of 4 with replacement and order does not
matter.
9* 8* 7
Answer: C(4+6-1,6) = C(9,6) =
= 84
3* 2*1
Selection without Replacement
Permutations (Order Matters): P(n,r)
Choose r times from a set of n items without replacement and order matters.
If we do not have replacement but order matters:
Then first drawing has n possibilities; the second drawing has (n-1), the third has
n!
(n-2), etc. By the product rule we have: n x (n-1) x (n-2) x ... x (n-r+1) =
.
(n  r)!
This is usually called permutations of n things taken r at a time. The number of
n!
possible outcomes is P(n,r) =
(n  r )!
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Examples
Example 1: Six CD's with 57 songs. CD player can be programmed to play any 20
songs in any order. How many ways can 20 different songs be played?
Method: Choose 20 from 57 without replacement, order matters
57!
Answer: P(57,20) =
37!
Example 2: Given n integers x 1x 2 …x n , all distinct. How many sorted
arrangements are possible (i.e. how many permutations of x 1… x n ?)
Method: Choose all n items from n, without replacement, order matters
n!
n!
Answer: P(n,n) =
=
= n!
(n  n)!
0!
Example 3: S = {a,b,c,d,e,f}
How many 3-letter words are in S * which have no character repeated?
Method: Choose 3 from 6 without replacement and order matters.
6!
Answer: P(6,3) =
= 6*5*4 = 120
3!
Combinations (Order Does Not Matter) without replacement: C(n,r)
Choose r times from a set of n items without replacement and order doesn't matter.
If order does not count and there is no replacement, then we must
determine how many of the permutations are the same objects but only differ in
order. That is: In how many orders can n distinct objects be arranged?
n!
P(n,n) =
= n!. Therefore, if order does not matter and we do not have
0!
P(n,r )
replacement, the number of possibilities for n objects taken r at a time is
=
P(r,r)
n!
. This is usually referred to as combinations of n things taken r at a time =
(n  r)!r!
C(n,r).
There is another important type of combination, combinations
with replacement. The sequence can repeat, but order does not
matter. (Discussed Earlier)
n!
Number of ways to choose r from n is C(n,r) =
(n  r )! r !
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Example
How many ways can you choose a committee of 5 from 7 people?
7!
7*6
Answer: C(n,r) =
=
= 21
2!5!
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Summary of Permutations and Combinations
Type (n things, r times)
Order
Replacement
Counts
Symbol
Expression
Permutations w/ repl.
Yes
Yes
None
nr
Permutations - no repl.
Yes
No
P(n,r)
n!
(n-r)!
Combinations - no repl.
No
No
C(n,r)
n!
(n-r)! r !
Combinations - w/ repl.
No
Yes
C(n+r-1,r)
(n+r-1)!
r ! (n-1)!
More Examples
Example 1: Place n indistinguishable objects into r distinguishable boxes. Number of
possible outcomes is C(n+r-1,r-1).
001000100010
(n+r-1) bits, (r-1) ones: C(n+r-1,r-1)
000110010000
Example 2: 12 identical flyers are to be placed into 4 mailboxes. How many ways
can this be done?
15 !
15 * 14 * 13
Answer: C(12+4-1,4-1) = C(15,3) =
=
= 455
12!3!
3* 2*1
What if each mailbox must receive at least 2 letters?
Method: Put 2 letters in every box. Then, how many ways are there to distribute
remaining 4 among 4 mailboxes?
7!
7* 6* 5
Answer: C(4+4-1,4-1) = C(7,3) =
=
= 35
4!3!
3* 2*1
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Example 3a: 15 basketball players are to be drafted by 3 professional teams
(A,B,C). Each team will draft 5 players. In how many ways can this be done?
Method: Choose 5 from 15 for A: C(15,5), 5 from remaining 10 for B: C(10,5), C has
remaining 5: C(5,5) = 1 (multiply all three together). This determines all the ways we
can partition the players where A chooses first, B second, and C last.
Answer: C(15,5)*C(10,5)
Example 3b: Now, if we cannot tell which team is which, we say the teams are
indistinguishable. If this is the case, how many ways can 15 players be partitioned
into 3 teams of 5 each? (i.e. - we want to know how many different sets {A,B,C}
there are. In 3a, we could have one division with
(A,B,C) = ({1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15}) and another with
(A,B,C) = ({6,7,8,9,10},{1,2,3,4,5},{11,12,13,14,15}).
But these two examples divide the fifteen players up into the same groups.)
C(15,5) *C(10,5)
Answer:
(since the teams are indistinguishable)
3!
Example 4: In how many ways can 2n people be divided into n pairs without
replacement and order does not matter?
C(2n, 2) * C(2n  2, 2) * ... * C(2, 2)
Answer:
n!
Example 5: In how many ways can 10 boys and 5 girls stand in a line if no 2 girls can
stand next to each other?
Boy1
Boy2
...
Boy10
Method: Choose one slot for each girl (Select 5 slots from 11, without replacement
and order matters.).
Answer: There are 10! ways to place the boys and P(11,5) ways to place the girls, so
by the product rule the answer is 10! * P(11,5).
Example 6: In how many ways can 2 distinct numbers be selected from 1,2,...,100 so
that their sum is even?
Answer: (2 Odds) C(50,2) + (2 Evens) C(50,2) = 50*49
Example 7: In how many ways can 2 distinct numbers be selected from 1,2,...,100 so
that their sum is odd?
Answer: Choose 1 even and 1 odd = C(50,1) *C(50,1)= 50*50
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CSC224/226 -Packet 6- Counting & Graph Theory
Pigeonhole Principle
If (k+1) or more objects are placed in k boxes, then there is at least one box
containing two or more objects.
Examples
Example 1: If you have a group of 367 people, there must be at least two people
who have the same birthday (there are only 366 possible birthdays).
Example 2: If you have a group of 27 English words, there must be at least two
words that begin with the same letter (there are only 26 possible letters in the English
language).
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Graph Information
I.
Basic Definitions
II.
Euler Circuit and Fleury's Algorithm
III.
Hamilton Cycle
IV.
Kruskal's Algorithm
V.
Matrix Representation of Graphs
VI.
Hasse Diagrams
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Graphs
Basic Definitions
path: (way of getting from one vertex to another via edges) - sequence of
vertices in which successive vertices are joined by an edge
length of path: number of edges in path
Example
paths: uvwxy (length 4)
uxy (length 2)
uxwustxzy (length 8)
closed path: first vertex = last vertex
cycle: closed path in which all vertices are distinct, except first = last
acyclic graph: graph with no cycle
Example
closed path: xyztx
closed path: swxtzyxs
closed path: uvwvu
acyclic graph:
also cycle
not cycle
not cycle
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A graph is connected if every pair of vertices is joined by a path.
Example
Connected
Not connected
The degree of a vertex is the number of edges touching the vertex. (loops count
twice)
Example
Vertex
u
v
x
Degree
2
2
4
y
z
4
4
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Euler Circuit and Fleury's Algorithm
Euler circuit: closed path which uses each edge exactly once
Can show there is an Euler circuit if and only if every vertex has even degree.
("easy" to find)
Fleury's algorithm will always find Euler circuit in connected graph if every vertex
has even degree.
Fleury's Algorithm
Start at any vertex v
While (still more edges incident with v) do
- if possible choose edge vw which does not disconnect
- else choose only vw and delete v
- delete edge vw
- v := w
end do
Example of Fleury's Algorithm
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etc.
Euler circuit: 23215341452
Hamilton Cycle
Hamilton cycle: cycle which uses each vertex once
"hard" problem (version of Traveling Salesman Problem)
Example
Hamilton cycle: uvtwxysu
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Kruskal's Algorithm
Minimum cost network to connect sites.
Given sites and link costs
Kruskal's Algorithm
Start with no edges.
Examine edges in non-decreasing order of weight.
If an edge does not create a cycle, add it.
Example of Kruskal's Algorithm
Matrix Representation of Graphs
Label vertices v 1,v 2 …v n
Adjacency matrix M: nxn matrix with M[x,y] = number of edges joining v x and v y .
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Examples of Adjacency Matrices
Graph isomorphism: Graphs G and H are isomorphic if there is some way to label
the vertices of H so that H is G (ie, has same adjacency matrix).
Examples
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Suppose M is the adjacency matrix of a graph G. Can prove by induction:
M [i,j] = number of length-k paths joining i and j
k
Reachability Matrix R
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Fourth Hour Exam Study Sheet
I.
Counting
A.
B.
C.
D.
II.
Basic Rules (Product, Union, Inclusion/Exclusion)
Selection with Replacement (Permutations, Combinations)
Selection without Replacement (Permutations, Combinations)
Pigeonhole Principle
Graphs
A.
Basic Definitions
B.
C.
D.
E.
F.
Euler Circuit and Fleury's Algorithm
Hamilton Cycle
Kruskal's Algorithm
Matrix Representation of Graphs
Hasse Diagrams
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