chapter 6 the principle of inclusion and exclusion

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Chapter 6 The principle of inclusion and exclusion

6.1 排容原則(the principle of inclusion and exclusion)

6.2 亂序 (derangements) 6.3 城堡多項式 (the rook polynomial)

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6.1 排容原則

Determine the number of positive integers n , 1n 30, that are divisible by 2, 3, or 5.

#(2|n) = 30/2=15

#(3|n) = 30/3=10

#(5|n) = 30/5=6

#(6|n) = 30/6=5

#(10|n) = 30/10=3

#(15|n) = 30/15=2

#(30|n) = 30/30=1

By 3

30 10,20

6,12,18,24

15 3,9,21,27

5,25

2,4,8,14,16,22,26,28

Ans: 15+10+6-5-3-2+1=22

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6.1 排容原則

例 1. (a) Find the number of integers between 1 and 250 that are not divisible by any of the integers 2, 3, 5, and 7. (b) the number of integers that not divisible by 2 nor by 7 but are divisible by 5

(a)

#(2|n)=125, #(3|n)=83, #(5|n)=50, #(7|n)=35

#(6|n)=41, #(10|n)=25, #(14|n)=17, #(15|n)=16, #(21|n)=11, #(35|n)=7

#(30|n)=8, #(42|n)=5, #(70|n)=3, #(105|n)=2,

#(210|n)=1

250-(125+83+50+35)+(41+25+17+16+11+7)-(8+5+3+2)+1=57

(b) 50-25-7+3= 21

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6.1 排容原則

例 2. Find the number of r-digit quaternary sequences in which each of the three digits 1, 2, and 3 appears at least once.

Let a1,a2,and a3 be the properties that the digits 1, 2, and 3 do not appear in a

sequence, respectively.

N(a1) = N(a2)= N(a3)=3r

N(a1a2) = N(a1a3)= N(a2a3)=2r

N(a1a2 a3) =1

N(a1’a2’ a3’)= 4r-(3r+3r+3r )+(2r+2r+2r)-1

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6.2 亂序 (derangements)

A permutation of these integers is said to be a derangement of the integers if no integer appears in its natural position.

Let ai be the property of a permutation in which the ith object is placed in its

forbidden position with i=1, 2, …, n.

N(ai)=(n-1)! 有

1

n個

N(aiaj) = (n-2)! 有

2

n個

N(aiaj ak) =(n-3)! 有

3

n個

…

N(a1a2… an)= 1 有

n

n個

dn=N(a’1a’2… a’n)= )()1()!2(2

)!1(1

! nnn

nn

nn

nn n

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6.2 亂序 (derangements)

Remark: dn 與 e-1 相關

!

1)1(

!2

1

!1

11

!!2!11

1

2

ne

n

xxxe

n

nx

dn n!e-1

dn 的遞迴關係

dn = (n-1) (dn-1 + dn-2 )

n2-n

122-n

3n2n

2n1n1nn

(-1)(-1)

)d2(d(-1)

)]2)d(n[-(d

]1)d(n[dndd

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6.2 亂序 (derangements)

例 1. Let n books be distributed to n children. The books are returned and distributed to the children again later on. In how many ways can the books be distributed so that no child will get the same book twice?

first time n! second time dn

total ways : n!dn(n!)2e-1

例 2. In how many ways can the integers 1, 2, 3, 4, 5, 6, 7, 8, and 9 be permuted such that no odd integer will be in its natural position?

!45

5!5

4

5!6

3

5!7

2

5!8

1

5!9

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6.2 亂序 (derangements)

例 3. Find the number of permutations of the letters a, b, c, d, e, and f in which neither the pattern ace nor the pattern fd appears.

例 4. In how many ways can the letters x, x, x, x, y, y, y, z, and z be arranged so that all the letters of the same kind are not in a single block?

!3)!4

!6

!3

!5

!2

!4()

!3!4

!8

!2!4

!7

!2!3

!6(

!2!3!4

!9

6!-4!-5!+3!

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6.3 城堡多項式 (the rook polynomial)

The problem of nontaking rooks: a rook is a chessboard piece which “captures” on both rows and column

Given a chessboard, let rk denote the

number of ways of placing k nontaking rooks on the board.r0 = 1

r1 = 4

r2 = 3

R(x) = 1 +4x+3x2

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6.3 城堡多項式 (the rook polynomial)

)(Cr)(Cr(C)r eki1kk

)CR(x,)CxR(x,C)R(x, ei

Expansion formula:

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6.3 城堡多項式 (the rook polynomial)

)CR(x,)CxR(x,C)R(x, ei

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6.3 城堡多項式 (the rook polynomial)

)CR(x,)CxR(x,C)R(x, ei (x+1)(1+2x)+x(1+x) = 1 +4x +3x2

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6.3 城堡多項式 (the rook polynomial)

Given a chessboard C, find the rook polynomial of C.

1+6x +10x2 +4x3