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Permutation

Example 1

Example 2Circular PermutationPermuting r of n objects

Example 1

Example 2

Example 3

Example 4Addition Rule

Example 1

Example 2Difference Rule

Example 1Inclusion Exclusion Rule

Example 1

Symmetric Relations

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PermutationsPermutations

Sanjay Jain, Lecturer, Sanjay Jain, Lecturer,

School of ComputingSchool of Computing

PermutationsPermutations

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

PermutationsPermutations

Theorem: Suppose a set A has n elements (where n1). 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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ExamplesExamples

Number of anagrams of SINGAPORE: This is same as premuting 9 distinct elements. 9!

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ExamplesExamples

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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ExamplesExamples

Letters of “SING” appear together, but not necessarily in that order.

T1: First permute “SING”, A, P, O, R, ET2: 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 PermutationsCircular Permutations

A

C B B A

C B

CA

A

B C

Circular PermutationsCircular 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)

)!1(!

nn

n

Convention 0!=1

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Permuting r of n objectsPermuting 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)

nPr

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Permuting r of n objectsPermuting 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 objectsPermuting r of n objectsProof:

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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ExampleExample

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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ExampleExampleSuppose 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 Bif 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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ExampleExample

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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ExampleExampleHow 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.

Total number of HC starting at a fixed vertex: 1*4*3*2*1=4!

T1: 1

T2: 4

T3: 3

T4: 2

T5: 1

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The Addition RuleThe 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 ij, Ai Aj = .

Then #(A) = #(A1) + #(A2) + …. + #(Ak).

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ExampleExample

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 dailyIn how many ways can one go from SIN to KL on a

particular day

Answer: 8+(2+2)+1 ways

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ExampleExample

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 waysT2: 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 RuleThe Difference Rule

Theorem: If A is a finite set and BA, then #(A - B) = #(A) - #(B).

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ExampleExample

How many three digit numbers have at least one digit repeated?

A---Set of three digit numbersB--- 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 RuleInclusion Exclusion Rule

Theorem: If A, B and C are finite sets, then#(AB) = #(A) + #(B) - #(AB)

#(ABC) = #(A) + #(B) + #(C) - #(AB) - #(AC) - #(BC) + #(ABC)

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ExampleExampleClass of 50 students.30 know Pascal18 know Fortran26 know Java9 know both Pascal and Fortran16 know both Pascal and Java8 know both Fortran and Java47 know at least one of the three languages.Question: How many know all three languages?

P: set of students who know PascalF: set of students who know FortranJ: set of students who know Java

ExampleExample

#(PFJ) = #(P) + #(F) + #(J) - #(PF) - #(PJ) -#(FJ) + #(PFJ)

47=30 + 18 + 26 - 9 - 16 - 8 + #(PFJ) #(PFJ) = 47 - 30 - 18 - 26 + 9 + 16 + 8

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