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Permutation
Example 1
Example 2
Circular Permutation
Permuting r of n objects
Example 1
Example 2
Example 3
Example 4
Addition Rule
Example 1
Example 2
Difference Rule
Example 1
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Permutations
Sanjay Jain, Lecturer,
School of Computing
Permutations
Permutation of a set of objects is an ordering of the
objects in a row.
Example: {A, B, C}
ABC, ACB, BAC, BCA, CAB, CBA
Permutations
Theorem: Suppose a set A has n elements (where
n1). Then the number of permutations of A is
n!= n*(n-1)*(n-2)*…*1.
Proof:
Job: select a permuation
T1: Select the 1st element in the row ---> n ways
T2: Select the 2nd element in the row ---> n-1 ways
……..
Tn: Select the nth element in the row ---> 1 way
Total number of permutations: n*(n-1)*…..*1
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Examples
Number of anagrams of SINGAPORE:
This is same as premuting 9 distinct elements.
9!
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Examples
Number of anagrams of SINGAPORE which have
“SING” as a substring:
We can think of “SING” as one element.
Thus there are a total of 6 elements to be permuted
(“SING”,A,P,O,R,E).
6!
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Examples
Letters of “SING” appear together, but not necessarily
in that order.
T1: First permute “SING”, A, P, O, R, E
T2: Permute letters of “SING”.
T1 can be done in 6! ways.
T2 can be done in 4! ways.
Total number of anagrams with the constraint: 6!*4!
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Circular Permutations
C
A
C
B
A
B
C
B
B
A
A
C
Circular Permutations
How many circular permutations are there?
Note that each circular permutation has n different row
permutations (by starting at different objects in the
circle)
n!
 (n  1)!
n
Convention 0!=1
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Permuting r of n objects
Suppose 1  r n.
An r-permutation of a set of n elements is an ordered
selection of r elements from the set.
The number of r-permutations of a set of n elements is
denoted by
P(n,r)
nP
r
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Permuting r of n objects
Theorem: Suppose n, r are integers with 1 r  n.
P(n,r) = n*(n-1)*…..(n-r+1)
= n!/(n-r)!
For r=0, we take P(n,0)=1.
Permuting r of n objects
Proof:
T1: Select the 1st element in the row
T2: Select the 2nd element in the row
……..
Tr: Select the rth element in the row
T1 can be done in n ways.
T2 can be done in n-1 ways.
…..
Tr can be done in n - (r - 1) = n - r + 1 ways.
Total number of r-permutations are:
n * (n - 1) * ….. * (n - r + 1)
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Example
Suppose there are 350 students. In how many ways
can one select president, secretary and treasurer if
no person can hold two posts?
Permuting 3 of 350 objects.
P(350,3)
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Example
Suppose A and B are finite sets.
How many different functions from A to B are 1--1?
A = {a1, a2,…, an}. B = {b1, b2,..., bm}
if n > m: No 1--1 functions from A to B
if n  m:
Want to select f(a1), f(a2),…, f(an) from the set B
All distinct.
Thus, we are finding a n-permutation from a set of m
objects.
P(m,n) ways.
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Example
How many bijective functions are there from A to B?
A = {a1, a2,…, an}. B = {b1, b2,..., bm}
If m  n: zero bijective functions.
If m=n:
Want to select f(a1), f(a2),…, f(an) from the set B
All distinct.
We are selecting n out of n objects.
P(n,n)=n!
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Example
How many Hamiltonian circuits are there in K5 ?
Assume that we start at a fixed vertex.
T1: Pick first vertex in HC (fixed to be v1 )
T2: Pick second vertex in HC.
T3: Pick third vertex in HC.
T4: Pick fourth vertex in HC.
T5: Pick fifth vertex in HC.
T1:
T2:
T3:
T4:
T5:
1
4
3
2
1
Total number of HC starting at a fixed vertex: 1*4*3*2*1=4!
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The Addition Rule
Theorem: Suppose a finite set A equals the union of k
distinct mutually disjoint sets A1, A2,…, Ak.
That is A= A1  A2  ….  Ak,
and, for ij, Ai Aj =  .
Then #(A) = #(A1) + #(A2) + …. + #(Ak).
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Example
Suppose I can go from SIN to KL by bus, train or plane.
There are
8 flights daily
2 morning and 2 evening trains, daily
1 bus daily
In how many ways can one go from SIN to KL on a
particular day
Answer:
8+(2+2)+1 ways
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Example
In Fortran identifiers consist of 1 to 6 characters where
the first character must be English letter and others
either English letter or a digit.
How many different identifiers are possible.
First step:
we calculate how many identifiers of length k are
there (where 1  k  6)
T1: Pick the first character
T2: Pick the second character
…..
Tk: Pick the kth character.
T1: 26 ways
T2: 36 ways
…..
Tk: 36 ways
26*36k-1 identifiers of length k
Second step:
Total number of identifiers=
26*360+26*361+ 26*362+... …..+26*365
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The Difference Rule
Theorem: If A is a finite set and BA, then
#(A - B) = #(A) - #(B).
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Example
How many three digit numbers have at least one digit
repeated?
A---Set of three digit numbers
B--- Set of three digit numbers which have no digit
repeated.
#(A) = 9*10*10
#(B) = 9*9*8
#(A - B) = 9*10*10 - 9*9*8
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Inclusion Exclusion Rule
Theorem: If A, B and C are finite sets, then
#(AB) = #(A) + #(B) - #(AB)
#(ABC) = #(A) + #(B) + #(C) - #(AB) - #(AC) #(BC) + #(ABC)
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Example
Class of 50 students.
30 know Pascal
18 know Fortran
26 know Java
9 know both Pascal and Fortran
16 know both Pascal and Java
8 know both Fortran and Java
47 know at least one of the three languages.
Question: How many know all three languages?
P: set of students who know Pascal
F: set of students who know Fortran
J: set of students who know Java
Example
#(PFJ) = #(P) + #(F) + #(J) - #(PF) - #(PJ) #(FJ) + #(PFJ)
47=30 + 18 + 26 - 9 - 16 - 8 + #(PFJ)
#(PFJ) = 47 - 30 - 18 - 26 + 9 + 16 + 8
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