Discrete Mathematics -- Chapter 8: The Principle of Ch t 8 Th P i - - PowerPoint PPT Presentation

Discrete Mathematics

Ch t 8 Th P i i l f

• - Chapter 8: The Principle of

Inclusion and Exclusion

Hung-Yu Kao (高宏宇) Department of Computer Science and Information Engineering, N l Ch K U National Cheng Kung University

Outline

The Principle of Inclusion and Exclusion Generalization of the Principle Generalization of the Principle Derangements: Nothing Is in Its Right Place Rook Polynomials Arrangements with Forbidden Positions Arrangements with Forbidden Positions

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8.1 The principle of Inclusion and Exclusion

For a given finite set S (|S| = N) with conditions C

) ( )] ( ) ( [ ) (

2 1 2 1 2 1

c c N c N c N N c c N + + − =

• For a given finite set S (|S| = N) with conditions Ci

) and (

2 1

c c N

) ( ) (

2 1 1

c c N c N − =

) ( ) ( ) (

2 1 2 1 2 1

c c N c c N N c c N ≠ − =

)

• r

( c c N

) (

2 1c

c N ≠

)

• r

(

2 1

c c N ) ( )] ( ) ( ) ( [ )] ( ) ( ) ( [ ) (

3 2 1 3 2 1

N N N N c N c N c N N c c c N + + − =

• Discrete Mathematics

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) ( )] ( ) ( ) ( [

3 2 1 3 2 3 1 2 1

c c c N c c N c c N c c N − + + +

Four sets

Ex 8.3 :

Four sets

)] ( ) ( ) ( ) ( ) ( ) ( [ )] ( ) ( ) ( ) ( [ ) (

4 3 4 2 3 2 4 1 3 1 2 1 4 3 2 1 4 3 2 1

c c N c c N c c N c c N c c N c c N c N c N c N c N N c c c c N + + + + + + + + + − =

For each element x∈S we have five cases:

) ( )] ( ) ( ) ( ) ( [ )] ( ) ( ) ( ) ( ) ( ) ( [

4 3 2 1 4 3 2 4 3 1 4 2 1 3 2 1 4 3 4 2 3 2 4 1 3 1 2 1

c c c c N c c c N c c c N c c c N c c c N c c N c c N c c N c c N c c N c c N + + + + − + + + + + +

For each element x∈S, we have five cases:

(0) x satisfies none of the four conditions; (1) x satisfies only one of the four conditions; (1) x satisfies only one of the four conditions; (2) x satisfies exactly two of the four conditions; (3) x satisfies exactly three of the four conditions;

(3) x satisfies exactly three of the four conditions;

(4) x satisfies all the four conditions.

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Four sets

) ( )] ( ) ( ) ( ) ( [ )] ( ) ( ) ( ) ( ) ( ) ( [ )] ( ) ( ) ( ) ( [ ) (

4 3 2 1 4 3 2 4 3 1 4 2 1 3 2 1 4 3 4 2 3 2 4 1 3 1 2 1 4 3 2 1 4 3 2 1

c c c c N c c c N c c c N c c c N c c c N c c N c c N c c N c c N c c N c c N c N c N c N c N N c c c c N + + + + − + + + + + + + + + − =

1.

x satisfies no condition. x is counted once in and once in N. [1=1]

2.

x satisfies c1. It is not counted on the left side. It is counted once in N and i ( ) [0 1 1 0]

) (

4 3 2 1

c c c c N

• nce in N(c1). [0 = 1-1 = 0]

3.

x satisfies c2 and c4. It is not counted on the left side. It is counted once in N, N(c2), N(c4) and N(c2c4).

( ) ( )

2 2

4.

x satisfies c1, c2 and c4. It is not counted on the left side. It is counted

• nce in N N(c1) N(c2) N(c4) N(c1c2) N(c1c4) N(c2c4) and N(c1c2c4)

( ) ( )

] 1 1 ) 1 1 ( 1 [

2 2 1 2

= + − = + + − =

• nce in N, N(c1), N(c2), N(c4), N(c1c2), N(c1c4), N(c2c4) and N(c1c2c4).

5.

x satisfies all conditions. It is not counted on the left side. It is counted

( ) ( ) ( )

] 1 1 ) 1 1 1 ( ) 1 1 1 ( 1 [

3 3 2 3 1 3

= − + − = − + + + + + − =

• nce in all the subsets on the right side.

( ) ( ) ( ) ( )

] 1 [

4 4 3 4 2 4 1 4

= + − + − =

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Four sets

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The Principle of Inclusion and Exclusion

Theorem 8.1:

|S| = N conditions c 1≤ i ≤ t |S| = N, conditions ci, 1≤ i ≤ t

• denote the number of elements of S

that satisfy none of the conditions

) (

2 1 t

c c c N N ⋅ ⋅ ⋅ =

that satisfy none of the conditions.

(2) ) ( ) ( ) (

1 1 1 t k j i k j i t j i j i t i i

c c c N c c N c N N N ⋅ ⋅ ⋅ + − + − =

∑ ∑ ∑

≤ < < ≤ ≤ < ≤ ≤ ≤

) ( ) 1 (

3 2 1 t t j j

c c c c N ⋅ ⋅ ⋅ − +

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The Principle of Inclusion and Exclusion

Corollary 8.1: Some notation for simplifying Theorem 8.1

. )

• r

...

• r
• r

(

2 1

N N c c c N

t

− =

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The Principle of Inclusion and Exclusion

Ex 8.4 : Determine the number of positive integers

n where 1 ≤ n ≤100 and n is not divisible by 2, 3 n where 1 ≤ n ≤100 and n is not divisible by 2, 3

• r 5.

Condition c if n is divisible by 2 Condition c1 if n is divisible by 2. Condition c2 if n is divisible by 3.

C di i if i di i ibl b 5

Condition c3 if n is divisible by 5. Then the answer to this problem is

⎣ ⎦

) ( )] ( ) ( ) ( [ )] ( ) ( ) ( [ ) (

3 2 1 3 2 3 1 2 1 3 2 1 3 2 1 3 2 1

− + + + + + − = − + − = c c c N c c N c c N c c N c N c N c N N S S S S c c c N

⎣ ⎦

16 ) 3 * 2 /( 100 =

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. 26 3 ] 6 10 16 [ ] 20 33 50 [ 100 = − + + + + + − =

The Principle of Inclusion and Exclusion

Ex 8.5 : Determine the number of nonnegative integer

solutions to the equation x1 + x2 + x3 + x4 = 18 and xi ≤ 7 for all i.

We say that a solution x1, x2, x3, x4 satisfies condition ci if xi > 7

(i.e., xi ≥ 8) .

Then the answer to this problem is

⎞ ⎛ 1 2 4

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⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + 18 1 18 4 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + 10 1 10 4

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + 2 1 2 4

The Principle of Inclusion and Exclusion

Ex 8.6 : For finite sets A, B, where ⎪A⎪= m ≥ n = ⎪B⎪, and function

f: A→B, determine the number of onto functions f.

L A { } d B {b b b }

• Let A = {a1, a2,…, am} and B = {b1, b2, …, bn}.
• Let ci be the condition that bi is not in the range of f.

range. in their have that in functions

• f

number the is ) ( Then

1 i

b S c N

• Then the answer to this problem is

). ...... (

2 1 n

c c c N

( ) ( )

m n m m n m i m

S N n S n c N n S S N ) 2 ( ) 2 ( ) ( ) 1 ( ) 1 ( ) ( | |

1 1

− = ⇒ − = = = =

( )

m n m j i

n S n c c N ) 2 ( ) 2 ( ) (

2 2

− = ⇒ − =

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The Principle of Inclusion and Exclusion

Ex 8.7 : In how many ways can the 26 letters of the

alphabet be permuted so that none of the patterns car, dog, pun or byte occurs? pun, or byte occurs?

Ex 8.8 : Let φ(n) be the number of positive integers m,

where 1 ≤ m < n and gcd(m, n)=1, i.e., m and n are relatively prime. relatively prime.

Consider For 1 ≤ i ≤ 4, let ci denote that k is divisible by pi.

S ( ) / ( ) /( )

4 3 2 1

4 3 2 1 e e e e

p p p p n =

N = S0 = n; N(ci) = n/pi; N(cicj) = n/(pipj); …

Then the answer to this problem is

). (

4 3 2 1

c c c c N

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1 4 4 1 3 3 1 2 2 1 1 1

4 3 2 1

) 1 ( ) 1 ( ) 1 ( ) 1 (

− − − −

− − − −

e e e e

p p p p p p p p

The Principle of Inclusion and Exclusion

Ex 8.9 : Six married couples are to be seated at a circular table. In how

many ways can they arrange themselves so that no wife sits next to her husband?

For 1≤ i ≤6, let ci denote the condition where a seating

arrangement has couple i seated next to each other. N(ci) = 2(11-1)! ( )

Then the answer to this problem is

). .... (

6 2 1

c c c N

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The Principle of Inclusion and Exclusion

8 10 i f h id fi ill

Ex 8.10 : In a certain area of the countryside are five villages.

An engineer is to devise a system of two-way roads so that after the system is completed, no village will be isolated. In y p g how many ways can he do this?

Let ci denote the condition that a system of these

roads isolates village a b c d and e respectively roads isolates village a, b, c, d, and e, respectively.

C(5,2) C(4,2)

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8.2 Generalizations of the Principle

• Em denotes the number of elements in S that satisfy exactly m of the t

conditions.

• )

( ) ( ) ( c c c c N c c c c N c c c c N E + + +

• ).

( ) ( ) ( ). ( ) ( ) (

1 2 2 1 3 2 1 3 2 1 2 1 2 1 3 2 1 3 2 1 1 t t t t t t t t t

c c c c c N c c c c N c c c c N E c c c c N c c c c N c c c c N E

− − −

⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ =

• E1= 2+3+4=N(c1)+N(c2)+N(c3) - 2[N(c1c2)+ N(c1c3)+ N(c2c3)] + 3N(c1c2c3)

= S1-2S2+3S3 =

( ) ( ) 3

2 3 2 1 2 1

S S S + −

• E2= 5+6+7=N(c1c2)+N(c1c3)+N(c2c3)-3N(c1c2c3)

= S2 - 3S3 =

• E =8= N(c c c ) = S

( ) 3

1 3 2

S S −

• E3=8= N(c1c2c3) = S3

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Generalizations of the Principle

( ) ( ) ( )

4 3 2 1 1

4 3 2 S S S S E − + − =

( ) ( ) ( ) 4

3 4 3 2 3 2 1 2 1

6 3 S S S E S S S S + − + − =

( ) ( ) 4

2 4 3 1 3 2 4 3 2 2

6 3 S S S S S S E + − = + − =

( ) ( ) ( ) 4

1 4 3 4 3 3 2 1

4 S S S S E + = − =

4 4

S E =

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Theorem 8.2

( ) ( ) ( )

(1) ) 1 (

2 1 t m t m m

S S S S E

− + +

( ) ( ) ( )

(1) . ) 1 (

2 2 2 1 1 1 t m t t m t m m m m m m

S S S S E

− − + + + +

− + ⋅ ⋅ ⋅ − + − =

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Corollary 8.2

Let Lm denotes the number of elements in S that satisfy at

least m of the t conditions.

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8.3 Derangements: Nothing Is in Its Right Place

• 1

⋅ ⋅ ⋅ + − + − = − = = ⋅ ⋅ ⋅ + + + + =

∑ ∑

∞ = − ∞ =

! 3 1 ! 2 1 1 1 ! ) 1 ( , ! ! 3 ! 2 1

1 3 2 n n n n x

n e n x x x x e

e-1=0.36788, 1-1+(1/2!)-(1/3!)+…-(1/7!) ≈ 0.36786

D t th t ll b i th iti

Derangement means that all numbers are in the wrong positions.

Ex 8.12 : Determine the number of derangements of 1, 2,…,10.

Let c be the condition that integer i is in the ith place Let ci be the condition that integer i is in the ith place.

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Derangements: Nothing Is in Its Right g g g Place

The general formula:

] 1 1 1 1 1 1 [ ! !

1

n e n d + + = =

−

P=d /n!

Ex 8 14 : Peggy has seven books and hires seven

] ! .... ! 4 ! 3 ! 2 1 1 [ ! ! n n e n dn − + − + − = =

P=dn/n!

Ex 8.14 : Peggy has seven books and hires seven

• reviewers. She wants two reviewers per book. In how

many ways can she make the distributions? many ways can she make the distributions?

The first time: 7! ways The second time: d7=7!* e-1Ways (different position)

7

y ( p )

Totally, we have 7!×d7 ways

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8.4 Rook Polynomials

In Fig. 8.6, we want to determine the number of ways in

which k rooks can be placed on the unshaded squares of this chessboard so that no two of them can take each

• ther—that is, no two of them are in the same row or

column of the chessboard C. This number is denoted as rk(C).

3 2 1 4 5 6

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Rook Polynomials

In Fig. 8.6, we have r0=1, r1= 6, r2= 8, r3= 2 and rk= 0 for

k ≥ 4.

Rook polynomial: r(C, x) = 1+6x+8x2+2x3. For each k ≥

0, the coefficient of xk is the number of ways we can place k nontaking rooks on chessboard C.

3 2 1 3 2 1 4 5 6

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Disjoint Subboards

Break up a larger board into smaller subboards. In Fig. 8.7, the chessboard contains two disjoint

g , j subboards that have no squares in the same column or row of C.

r(C, x) = r(C1, x) . r(C2, x)

2 4 1 ) (

2

C

C1

2 10 7 1 ) , ( 2 4 1 ) , (

3 2 2 2 1

x x x x C r x x x C r + + + = + + =

C2

) , ( ) , ( 4 28 56 40 11 1 ) , (

2 1 5 4 3 2

x C r x C r x x x x x x C r ⋅ = + + + + + =

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Disjoint Subboards

f C

r3 for C In general, if C is a chessboard made up of pairwise

disjoint subboards C1, C2,…, Cn, then r(C, x) = r(C1, x). j

1 2 n

( ) (

1

) r(C2, x)…. r(Cn, x).

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Recursive Formula

• Fig. 8.8 (a), For a given designated square (*), we either (b) place one

rook here, or (c) do not use this square.

r (C) = r

(C )+r (C )

rk(C) = rk-1(Cs)+rk(Ce)

• Cs: denote the remaining smaller subboard (Fig. 8.8(b))
• Ce: C with the one designed square eliminated (Fig. 8.8(c))

e

g q ( g ( ))

rk(C)xk = rk-1(Cs) xk + rk(Ce)xk for 1≤ k ≤ n.

use not use

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Recursive Formula

r0(Cs)x0 typo in p405

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Apply the Recursive Formula

use * not use * use * not use *

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8.5 Arrangements with Forbidden Positions

• Ex 8.15 : In making seating arrangements, the shaded square of the figure

means relative Ri will not sit at table Tj.

Determine the number of ways that we can seat these four

y relatives at five different tables.

Let |S| be the total number of ways we can place the four

• relatives. (|S| = 5!)

(| | )

Let ci be the condition that

Ri is seated in a forbidden position but at different tables.

T1 T2 T3 T4 T5 R1

p

• N(c1) = 4! + 4! (R1→ T1 or R1→ T2)
• N(c2) = 4! (R2→ T2)
• N(c3) = ?

1

R2 R

4! + 4!

• N(c4) = ?

S1 = 7(4!)

R3 R4

4! + 4!

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number of shaded squares

Arrangements with Forbidden Positions

• S2 = 16(3!)

N(c1c2) = 3! (R1→ T1 or R2→ T2) N(c1c3) = 4(3!) N(c1c4) = ?, N(c2c3) = ?, N(c2c4) = ?, N(c3c4) = ?

Observation: 16 is the number of ways two nontaking rooks can be

placed on the shaded chessboard.

Let ri be the number of ways in which it is possible to place i nontaking

rooks on the shaded chessboard.

• For all 0 ≤ i ≤ 4, Si = ri(5 - i)!

i i

Decompose C into the disjoint subboards in the upper left and lower

right corners.

• r(C, x)=(1 + 3x + x2)(1 + 4x + 3x2) = 1 + 7x + 16x2 + 13x3 + 3x4

( , ) ( )( )

2 )! ( ) 1 ( ) ! 1 ( 3 ) ! 2 ( 13 ) ! 3 ( 16 ) ! 4 ( 7 ) ! 5 ( 1 ) (

4 4 3 2 1 4 3 2 1

+ − + − = + − + − = ∴

∑

i

S S S S S c c c c N

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25 )! 5 ( ) 1 ( = − − =∑

= i i i

i r

Arrangements with Forbidden Positions

E 8 16 W ll di i i h i d di d h h

• Ex 8.16 : We roll two dice six times, where one is red die and the other green

die.

• We know the following pairs did not occur: (1, 2), (2, 1), (2, 5), (3, 4), (4, 1),

(4, 5) and (6, 6).

• What is the probability that we obtain all six values both on red die and green

die?

One of solutions is like (1, 1), (2, 3), (4, 4), (3, 2), (5, 6), (6, 5).

• In Fig. 8.10(b), chessboard C with seven shaded squares

(C ) (1+4 +2 2)(1+ )3 1+7 +17 2+19 3+10 4+2 5

r(C, x) = (1+4x+2x2)(1+x)3 = 1+7x+17x2+19x3+10x4+2x5 ci denotes that all six values occur on both the red and green dies,

but i on the red die is paired with one of the forbidden numbers on p the green die.

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Figure 8 10

1 2 3 4 5 6 1 5 3 4 2 6

(b) Figure 8.10 (a)

1 2 1 2 2 3 2 4 4 5 3 5 6 6

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Arrangements with Forbidden Positions Arrangements with Forbidden Positions

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Arrangements with Forbidden Positions

B

Ex 8.17 : How many

• ne-to-one functions f: A → B

ti f f th f ll i

u v w x y z 1

B

satisfy none of the following conditions:

1 2 3

w f c v u f c ) 2 ( :

• r

) 1 ( :

2 1

= =

A

4

z y x f c x w f c f

• r

, , ) 4 ( :

• r

) 3 ( : ) (

4 3 2

= =

r(C, x) = (1+2x)(1+6x+9x2+2x3) = 1+8x+21x2+20x3+4x4

• P(6,4)

P(5,3)

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Arrangements with Forbidden Positions

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8

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Exercises

You should explain your answers in

detail, only one-line answer will be rated detail, only one line answer will be rated

8-1: 6, 13, 16

8-2: 5 8-2: 5 8-3: 6, 10 8-4,5: 5, 12 (add f(5)≠z)

How many integers n are such that 0≤n<10,000 and the sum of

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y g , the digits is less than or equal to 27?