1 Inclusion-Exclusion Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong.

1

Inclusion-Exclusion

Supplementary Notes

Prepared by Raymond WongPresented by Raymond Wong

2

e.g.1 (Page 3)

What is the size of E, denoted by S(E)?

E

b

c

F

a

de

f

What is the size of F, denoted by S(F)?

What is the size of E U F, denoted by S(E U F)?

3

4

6

Please express S(E U F) in terms of S(E), S(F) and S(E F).

What is the size of E F, denoted by S(E F)? 1

S(E U F) = S(E) + S(F) – S(E F)

3

e.g.2 (Page 5)

We know that S(E U F) = S(E) + S(F) – S(E F)

where S(E) is the size of E

E

b

c

F

a

de

f

This principle also applies in probabilities Let E and F be two events. We have

P(E U F) = P(E) + P(F) – P(E F) where P(E) is the probability of event E

4

e.g.3 (Page 6) Consider we roll two dice. Let E be the event that the sum of the two

dice is even Let F be the event that the sum of the two

dice is 8 or more.

E: sum is evenF: sum is 8 or more

What is P(E)? What is P(F)? What is P(E F)? What is P(E U F)?

What is P(E)?What is P(F)?What is P(E F)?What is P(E U F)?

5

e.g.3



Dice 1

Dice 2

Sum

1 1 2

1 2 3

1 3 4

1 4 5

1 5 6

1 6 7

2 1 3

2 2 4

2 3 5

2 4 6

2 5 7

2 6 8

Dice 1

Dice 2

Sum

3 1 4

3 2 5

3 3 6

3 4 7

3 5 8

3 6 9

4 1 5

4 2 6

4 3 7

4 4 8

4 5 9

4 6 10

Dice 1

Dice 2

Sum

5 1 6

5 2 7

5 3 8

5 4 9

5 5 10

5 6 11

6 1 7

6 2 8

6 3 9

6 4 10

6 5 11

6 6 12

We want to find P(sum is even)= 1/2

1/2

6

e.g.3



Dice 1

Dice 2

Sum

1 1 2

1 2 3

1 3 4

1 4 5

1 5 6

1 6 7

2 1 3

2 2 4

2 3 5

2 4 6

2 5 7

2 6 8

Dice 1

Dice 2

Sum

3 1 4

3 2 5

3 3 6

3 4 7

3 5 8

3 6 9

4 1 5

4 2 6

4 3 7

4 4 8

4 5 9

4 6 10

Dice 1

Dice 2

Sum

5 1 6

5 2 7

5 3 8

5 4 9

5 5 10

5 6 11

6 1 7

6 2 8

6 3 9

6 4 10

6 5 11

6 6 12

We want to find P(sum is 8 or more)

1/2

P(sum is 8)

P(sum is 9)

P(sum is 10)

P(sum is 11)

P(sum is 12)

= 5/36

= 4/36

= 3/36

= 2/36

= 1/36

= 5/36 + 4/36 + 3/36 + 2/36 + 1/36= 15/36

15/36

7

e.g.3



Dice 1

Dice 2

Sum

1 1 2

1 2 3

1 3 4

1 4 5

1 5 6

1 6 7

2 1 3

2 2 4

2 3 5

2 4 6

2 5 7

2 6 8

Dice 1

Dice 2

Sum

3 1 4

3 2 5

3 3 6

3 4 7

3 5 8

3 6 9

4 1 5

4 2 6

4 3 7

4 4 8

4 5 9

4 6 10

Dice 1

Dice 2

Sum

5 1 6

5 2 7

5 3 8

5 4 9

5 5 10

5 6 11

6 1 7

6 2 8

6 3 9

6 4 10

6 5 11

6 6 12

We want to find P(even sum is 8 or more)

1/2

P(sum is 8)

P(sum is 10)

P(sum is 12)

= 5/36

= 3/36

= 1/36

= 5/36 + 3/36 + 1/36 = 9/36

15/369/36

8

e.g.3



We want to find P(E U F)

1/215/369/36

(By the Principle of Inclusion and Exclusion)

P(E U F)

= P(E) + P(F) – P(E F)

= 1/2 + 15/36 – 9/36

= 2/3

2/3

9

e.g.4 (Page 6)


E

b

c

F

a de

f


What is the size of E U F U G, denoted by S(E U F U G)?

5

6

11 Is “S(E U F U G) = S(E) + S(F) + S(G)

- S(E F) – S(E G) – S(F G)”?

G

gh i

j

k

What is the size of G, denoted by S(G)? 5

What is the size of E F, denoted by S(E F)? 2 What is the size of E G, denoted by S(E G)? 2

What is the size of F G, denoted by S(F G)? 2 What is the size of EFG, denoted by S(EFG)? 1

No

RHS = 5 + 6 + 5 – 2 – 2 – 2= 16 – 6 = 10LHS = 11

10

e.g.4


E

b

c

F

a de

f


What is the size of E U F U G, denoted by S(E U F U G)?

5

6

11 “S(E U F U G) = S(E) + S(F) + S(G)

- S(E F) – S(E G) – S(F G) + S(EFG)”

G

gh i

j

k

What is the size of G, denoted by S(G)? 5

What is the size of E F, denoted by S(E F)? 2 What is the size of E G, denoted by S(E G)? 2

What is the size of F G, denoted by S(F G)? 2 What is the size of EFG, denoted by S(EFG)? 1

RHS = 5 + 6 + 5 – 2 – 2 – 2 + 1= 16 – 6 + 1= 11LHS = 11

11

e.g.5 (Page 6)

We know that S(E U F U G) = S(E) + S(F) + S(G) - S(E F) – S(E G) – S(F G) + S(EFG)

where S(E) is the size of E This principle also applies in probabilities Let E, F and G be three events. We have

P(E U F U G) = P(E) + P(F) + P(G) - P(E F) – P(E G) – P(F G) + P(EFG)

where P(E) is the probability of event E

12

e.g.6 (Page 10) We have seen

P(E U F) = P(E) + P(F) – P(E F)

We re-write asP(E1 U E2) = P(E1) + P(E2) – P(E1 E2)

We further re-write as

)()( 21

2

1

2

1

EEPEPEPi

ii

i

13

e.g.7 (Page 10) We have seen

P(E U F U G) = P(E) + P(F) + P(G) - P(E F) – P(E G) – P(F G) + P(EFG)

We re-write asP(E1 U E2 U E3) = P(E1) + P(E2) + P(E3) - P(E1 E2) – P(E1 E3) – P(E2 E3) + P(E1E2E3)


)()()( 321

2

1

3

1

3

1

3

1

EEEPEEPEPEPi ij

jii

ii

i


31:,,

31:,

31:

3

1

321

321

321

21

21

21

1

1

1)()()(

iiiiii

iii

iiii

ii

ii

ii

i EEEPEEPEPEP

14

e.g.7

From

31:,,

31:,

31:

3

1

321

321

321

21

21

21

1

1

1)()()(

iiiiii

iii

iiii

ii

ii

ii

i EEEPEEPEPEP

we further re-write as

31:,,

13

31:,

12

31:

11

3

1

321

321

321

21

21

21

1

1

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

iiiiii

iii

iiii

ii

ii

i

ii

EEEPEEPEP

EP

15

e.g.7

31:,,

13

31:,

12

31:

11

3

1

321

321

321

21

21

21

1

1

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

iiiiii

iii

iiii

ii

ii

i

ii

EEEPEEPEP

EP

16

e.g.7

31:,,

13

31:,

12

31:

11

3

1

321

321

321

21

21

21

1

1

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

iiiiii

iii

iiii

ii

ii

i

ii

EEEPEEPEP

EP

we can re-write as

3...1:,...,,

3

1

13

121

21

21)...()1(

k

k

k

iiiiii

iiik

k

ii EEEPEP

17

e.g.7

3...1:,...,,

3

1

13

121

21

21)...()1(

k

k

k

iiiiii

iiik

k

ii EEEPEP

18

e.g.8 (Page 13)

3...1:,...,,

3

1

13

121

21

21)...()1(

k

k

k

iiiiii

iiik

k

ii EEEPEP

According to

we deduce a general formula as follows

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Prove that

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Why is it correct?

19

e.g.8

Prove that

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Step 1: Prove that P(2) (i.e., the base case) is true.

21

:,

12

21:

11

21

21

21

1

1

1)()1()()1(

iiii

ii

ii

i EEPEP

We want to show that

2...1:,...,,

2

1

12

121

21

21)...()1(

k

k

k

iiiiii

iiik

k

ii EEEPEP

(*)

RHS of (*)

2...1:,...,,

2

1

1

21

21

21)...()1(

k

k

k

iiiiii

iiik

k EEEP

)()1())()(()1( 2112

2111 EEPEPEP

)()()( 2121 EEPEPEP In some slides, we know that P(E1 U E2) = P(E1) + P(E2) – P(E1 E2)

Thus, P(2) is true.

20

e.g.8

Prove that

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Step 2: Prove that “P(n-1) P(n)” is true for all n > 2.

Step 2(a): Assume that P(n-1) is true for n > 2.

That is,

1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

In other words, for any sets F1, F2, …, Fn-1, we have

1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

21

e.g.8

Prove that

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(



Step 2(b): According to P(n-1), we deduce that P(n) is true.


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

We want to show that

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Objective:

22

e.g.8

Prove that

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(



Step 2(b): According to P(n-1), we deduce that P(n) is true.


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

Objective:

Consider

n

iiEP

1

P(E U F) = P(E) + P(F) – P(E F)(proved in the base case)

n

n

ii EEP )(

1

1

n

n

iin

n

ii EEPEPEP )(

1

1

1

1

23

e.g.8Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

Objective:

Consider

n

iiEP

1

n

n

iin

n

ii EEPEPEP )(

1

1

1

1

Inductive Hypothesis:

24

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

Objective:

Consider

n

iiEP

1

n

n

iin

n

ii EEPEPEP )(

1

1

1

1


nnn

n

ii EEEEPEPEP

)...( 121

1

1

)(...)()( 121

1

1nnnnn

n

ii EEEEEEPEPEP

(By Distributive Law)

1

1

1

1

)(n

inin

n

ii EEPEPEP

Let Gi = Ei En for i < n

1

1

1

1

n

iin

n

ii GPEPEP

n

iiEP

1

25

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

Objective:

Consider

n

iiEP

1



1

1

1

1

n

iin

n

ii GPEPEP

n

iiEP

1

26

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

Objective:

Consider

n

iiEP

1



1

1

1

1

n

iin

n

ii GPEPEP

nniii

iiiiii

n

k

k EPEEEP

k

k

k

1...1:,...,,

1

1

1

21

21

21)...()1(

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

iii

n

k

k

k

k

kGGGP

)(...)()(21 ninini EEEEEE

kninini EEEEEE

k ...

21

niii EEEEk ...

21

kiii GGG ...21

Consider

kiii GGG ...21

niii EEEEk ...

21

27

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(


1...1:,...,,

1

1

11

121

21

21)...()1(

niiiiii

iii

n

k

kn

ii

k

k

kFFFPFP

Objective:

Consider

n

iiEP

1



1

1

1

1

n

iin

n

ii GPEPEP

nniii

iiiiii

n

k

k EPEEEP

k

k

k

1...1:,...,,

1

1

1

21

21

21)...()1(

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

iii

n

k

k

k

k

kGGGP

kiii GGG ...21

niii EEEEk ...

21

nniii

iiiiii

n

k

k EPEEEP

k

k

k

1...1:,...,,

1

1

1

21

21

21)...()1(

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

niii

n

k

k

k

k

kEEEEP

28

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(Objective:

Consider

n

iiEP

1

nniii

iiiiii

n

k

k EPEEEP

k

k

k

1...1:,...,,

1

1

1

21

21

21)...()1(

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

niii

n

k

k

k

k

kEEEEP

29

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21


Consider

n

iiEP

1

nniii

iiiiii

n

k

k EPEEEP

k

k

k

1...1:,...,,

1

1

1

21

21

21)...()1(

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

niii

n

k

k

k

k

kEEEEP

nniii

iiiiii

n

k

k EPEEEP

k

k

k

1...1:,...,,

1

1

1

21

21

21)...()1(

1...1:,...,,

1

1

2

21

21

21)...()1(

niiiiii

niii

n

k

k

k

k

kEEEEP

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEPConsider

…………(**)

30

e.g.8

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEPConsider

31

e.g.8

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEPConsider

niniiii

iiiiiiii

n

k

kn

k

kk

kk

kkEEEEPEP

1

121

121

121

and...1

:,,...,,

1

1

2 )...()1()(

niniiii

iiiiiiii

n

a

an

a

aa

aa

aaEEEEPEP

and...1

:,,...,,2

1

121

121

121)...()1()(

(where k+1 = a)

niniiii

iiiiiiii

n

k

kn

k

kk

kk

kkEEEEPEP

and...1

:,,...,,2

111

121

121

121)...()1()()1(

niniiii

iiiiiiii

n

k

k

k

kk

kk

kkEEEEP

and...1

:,,...,,1

1

121

121

121)...()1(

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

121 nn

niniiii

iiiiiiii

n

k

k EEEPEEEEP

k

kk

kk

kk

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEP

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

32

e.g.8

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEP

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

33

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21


From (**), we have

n

iiEP

1

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

iii

n

k

k

k

k

kEEEP

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEP

1...1:,...,,

1

1

2

21

21

21)...()1()(

niiiiii

niii

n

k

kn

k

k

kEEEEPEP

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

iii

n

k

k

k

k

kEEEP

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

34

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21


From (**), we have

n

iiEP

1

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

iii

n

k

k

k

k

kEEEP

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

35

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21


From (**), we have

n

iiEP

1

1...1:,...,,

1

1

1

21

21

21)...()1(

niiiiii

iii

n

k

k

k

k

kEEEP

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

niniii

iiiiii

n

k

k

k

k

k

kEEEP

and...1

:,...,,

1

1

1

21

21

21)...()1(

)...()1()...()1( 211

and...1

:,,...,,

1

1

1

121

121

21 nn

niniiii

iiiiiii

n

k

k EEEPEEEP

k

kk

kk

k

)...()1()...()1( 211

...1:,...,,

1

1

1

21

21

21 nn

niiiiii

iii

n

k

k EEEPEEEP

k

k

k

niiiiii

iii

n

k

k

k

k

kEEEP

...1:,...,,1

1

21

21

21)...()1(

Thus, P(n) is true.

36

e.g.8

Let P(n) be

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21


We prove that “P(n-1) P(n)” is true for all n > 2

By Mathematical Induction, n 2,

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

37

e.g.9 (Page 22)

We know that What is P(E1 U E2 U E3 U E4)?

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

P(E1 U E2 U E3 U E4)= P(E1) + P(E2) + P(E3) + P(E4)

- P(E1 E2) - P(E1 E3) - P(E1 E4) - P(E2 E3) - P(E2 E4) - P(E3 E4)

+ P(E1 E2 E3) + P(E1 E2 E4) + P(E1 E3 E4) + P(E2 E3 E4)

- P(E1 E2 E3 E4)

38

e.g.10 (Page 23) There are 5 students who have the same model

and color of backpack. They put their backpacks randomly along the wall. Someone mixed up the backpacks so students get

back “random” backpacks. Suppose that there are two students called

“Raymond” and “Peter” (a) What is the probability that Raymond gets his

OWN backpack back? (b) What is the probability that Raymond and

Peter get their OWN backpacks back?

(a) What is the probability that Raymond gets his OWN backpack back? (b) What is the probability that Raymond and Peter get their OWN backpacks back?

39

e.g.10


Raymond Peter

(a)

There are (5-1)! cases that Raymond gets his OWN backpack back.

There are totally 5! cases

P(Raymond gets his OWN backpack back) = (5-1)!

5!

40

e.g.10


Raymond Peter

(b)

There are (5-2)! cases that Raymond and Peter get their OWN backpacks back.

There are totally 5! cases

P(Raymond and Peter get their OWN backpacks back) = (5-2)!

5!

41

e.g.11 (Page 23) There are n students who have the same model

and color of backpack. They put their backpacks randomly along the wall. Someone mixed up the backpacks so students get

back “random” backpacks. Suppose that there are two students called

“Raymond” and “Peter” (a) What is the probability 1 specified student

gets his OWN backpack back? (b) What is the probability k specified students

get their OWN backpacks back?

(a)What is the probability 1 specified student gets his OWN backpack back?

(b) What is the probability k specified students get their OWN backpacks back?

42

e.g.11

Raymond Peter

…

n

There are (n-1)! cases that 1 specified student gets his OWN backpack back.

There are totally n! cases

P(1 specified student gets his OWN backpack back) = (n-1)!

n!



(a)

= (n-1)!

(n-1)!.n=

1

n

1

n

43

e.g.11

Raymond Peter

…

n

…

k

There are (n-k)! cases that k specified students get their OWN backpacks back.

There are totally n! cases

P(k specified students their OWN backpacks back) = (n-k)!

n!



(b)

1

n(n-k)!

n!

44

e.g.12 (Page 26) Suppose that there are 5 students (i.e., n =

5) Let Ei be the event that student i gets his

own backpack back. What is the probability that at least one

person gets his own backpack?



1

n(n-k)!

n!

5...1:,...,,

5

1

1

21

21

21)...()1(

k

k

k

iiiiii

iiik

k EEEP

P(at least one person gets his own backpack)

= P(E1 U E2 U E3 U E4 U E5)

45

e.g.12



1

n(n-k)!

n!

5...1:,...,,

5

1

1

21

21

21)...()1(

k

k

k

iiiiii

iiik

k EEEP


= P(E1 U E2 U E3 U E4 U E5)

5...1:,...,,

5

1

1

21

21!5

)!5()1(

k

kiiiiiik

k k

P(k specified students get their own backpacks back)

Let Ei be the event that student i gets his own backpack back.

How many possible tuples in form of (i1, i2, …, ik) where 1 i1 < i2 < … < ik 5?5

k!5

)!5(5)1(

5

1

1 k

kk

k

!5

)!5(

)!5(!

!5)1(

5

1

1 k

kkk

k

!

1)1(

5

1

1

kk

k

46

e.g.12



1

n(n-k)!

n!



!

1)1(

5

1

1

kk

k

47

e.g.12



1

n(n-k)!

n!



!

1)1(

5

1

1

kk

k

!5

1)1(

!4

1)1(

!3

1)1(

!2

1)1(

!1

1)1( 1514131211

P(at least one person gets his own backpack)!5

1

!4

1

!3

1

!2

11

!5

1

!4

1

!3

1

!2

11

48

e.g.13 (Page 28) Suppose that there are 5 students (i.e., n =

5) Let Ei be the event that student i gets his

own backpack back. What is the probability that nobody gets

his own backpack?



1

n(n-k)!

n!

P(nobody gets his own backpack)

= 1 – P(at least one person gets his own backpack)

P(at least one person gets his own backpack)!5

1

!4

1

!3

1

!2

11

)!5

1

!4

1

!3

1

!2

11(1

!5

1

!4

1

!3

1

!2

1

49

e.g.14 (Page 29) Suppose that there are n students Let Ei be the event that student i gets his

own backpack back. What is the probability that at least one

person gets his own backpack?



1

n(n-k)!

n!

niiiiii

iii

n

k

k

k

k

kEEEP

...1:,...,,1

1

21

21

21)...()1(


= P(E1 U E2 U … U En)

50

e.g.14



1

n(n-k)!

n!

niiiiii

iii

n

k

k

k

k

kEEEP

...1:,...,,1

1

21

21

21)...()1(


= P(E1 U E2 U … U En)

niiiiii

n

k

k

k

kn

kn

...1:,...,,1

1

21

21!

)!()1(

P(k specified students get their own backpacks back)


How many possible tuples in form of (i1, i2, …, ik) where 1 i1 < i2 < … < ik n?n

k!

)!()1(

1

1

n

kn

k

nn

k

k

!

)!(

)!(!

!)1(

1

1

n

kn

knk

nn

k

k

!

1)1(

1

1

k

n

k

k

51

e.g.12



1

n(n-k)!

n!



!

1)1(

1

1

k

n

k

k

52

e.g.12



1

n(n-k)!

n!



!

1)1(

1

1

k

n

k

k

P(at least one person gets his own backpack)!

1)1(

1

1

k

n

k

k

53

e.g.15 (Page 29) Suppose that there are n students Let Ei be the event that student i gets his

own backpack back. What is the probability that nobody gets

his own backpack?



1

n(n-k)!

n!

P(nobody gets his own backpack)


P(at least one person gets his own backpack)!

1)1(

1

1

k

n

k

k

)!

1)1((1

1

1

k

n

k

k

Dearrangement Problem

54

e.g.15P(nobody gets his own backpack)


)!

1)1((1

1

1

k

n

k

k

)!

1)1(...

!3

1)1(

!2

1)1(

!1

1)1((1 1131211

nn

)!

1)1(...

!3

1

!2

1

!1

1(1 1

nn

!

1)1(...

!3

1

!2

1

!1

11 2

nn

!

1)1(...

!3

1

!2

1

!1

11

nn

!

)1(...

!3

)1(

!2

)1()1(1

32

n

n

Note that from calculus, we have

0

32

!...

!3!21

i

ix

n

xxxxe

1e if n is a large number

55



!

)1(...

!3

)1(

!2

)1()1(1

32

n

n

56



!

)1(...

!3

)1(

!2

)1()1(1

32

n

n

n !

)1(...

!3

)1(

!2

)1()1(1

32

n

n

e-1 = 0.367879441

57

e.g.16 (Page 33) Principle of Inclusion and Exclusion for Probability

niiiiii

iii

n

k

kn

ii

k

k

kEEEPEP

...1:,...,,1

1

121

21

21)...()1(

Principle of Inclusion and Exclusion for Counting

niiiiii

iii

n

k

kn

ii

k

k

kEEEE

...1:,...,,1

1

121

21

21...)1(

58

e.g.17 (Page 33) How many functions from a 6-

element set N to a 5-element set M = {y1, y2, …, y5} are there?

NM

1

2y1

y2

5 choices

5 choices

Total no. of functions = 5 x 5 x 5 x 5 x 5 x 5= 56

35 choices

4

5

6

5 choices

5 choices

5 choices

y3

y4

y5

59


element set N to a 5-element set M = {y1, y2, …, y5} map nothing to y1?

NM

1

2y1

y2

4 choices

4 choices


34 choices

4

5

6

4 choices

4 choices

4 choices

y3

y4

y5

60


element set N to a 5-element set M = {y1, y2, …, y5} map nothing to y1 and y2?

NM

1

2y1

y2

3 choices

3 choices


33 choices

4

5

6

3 choices

3 choices

3 choices

y3

y4

y5

61

e.g.20 (Page 33) How many functions from a 6-element

set N to a 5-element set M = {y1, y2, …, y5} map nothing to a given set K of k elements in M (e.g., {y1, y2})?N

M1

2y1

y2

(5-k) choices

Total no. of functions = (5-k) x (5-k) x (5-k) x (5-k) x (5-k) x (5-k) = (5-k)6

34

5

6

y3

y4

y5

(5-k) choices

(5-k) choices

(5-k) choices

(5-k) choices

(5-k) choices

Total no. of functions that map nothing to a given set K of k elements in M= (5-k)6

62

e.g.21 (Page 34) How many functions from a 6-element

set N to a 5-element set M = {y1, y2, …, y5} map nothing to at least one element in M?


Let Ei be a set of functions which map nothing to element yi

Total no. of functions that map nothing to at least one element in M

= E1 U E2 U … U E5

5...1:,...,,

5

1

1

21

21

21...)1(

k

k

k

iiiiii

iiik

k EEE

NM

1

2y1

y2

34

5

6

y3

y4

y5

5

1iiE=

niiiiii

iii

n

k

kn

ii

k

k

kEEEE

...1:,...,,1

1

121

21

21...)1(Principle of Inclusion-and-Exclusion

63

e.g.21 How many functions from a 6-element





5...1:,...,,

5

1

1

21

21

21...)1(

k

k

k

iiiiii

iiik

k EEE

NM

1

2y1

y2

34

5

6

y3

y4

y5

64

e.g.21 How many functions from a 6-element





5...1:,...,,

5

1

1

21

21

21...)1(

k

k

k

iiiiii

iiik

k EEE Total no. of functions that map nothing to a given set K of k elements where K = {yi1, yi2, …, yik}

5...1:,...,,

65

1

1

21

21

)5()1(

k

kiiiiiik

k k How many possible tuples in form of (i1, i2, …, ik) where 1 i1 < i2 < … < ik 5?5

k

65

1

1 )5(5

)1( kkk

k

5

1

61 )5(5

)1(k

k kk

NM

1

2y1

y2

34

5

6

y3

y4

y5

Total no. of functions that map nothing to at least one element in M=

5

1

61 )5(5

)1(k

k kk

65

e.g.22 (Page 37) How many onto functions from a 6-element

set N to a 5-element set M = {y1, y2, …, y5} are there?


5

1

61 )5(5

)1(k

k kk

NM

1

2y1

y2

34

5

6

y3

y4

y5

66

e.g.22

Onto function (or surjection)N M

N M


5

1

61 )5(5

)1(k

k kk

67

e.g.22

Not onto function (or not surjection)N M

NS M


5

1

61 )5(5

)1(k

k kk

68

e.g.22 How many onto functions from a 6-element

set N to a 5-element set M = {y1, y2, …, y5} are there?


5

1

61 )5(5

)1(k

k kk

Total no. of onto functions from a 6-element set N to a 5-element set M= Total no. of functions from a 6-element set N to a 5-element set M - Total no. of functions that are NOT onto

NM

1

2y1

y2

34

5

6

y3

y4

y5

From “e.g.,17”, Total no. of functions from a 6-element set N to a 5-element set M = 56

= 56 -

5

1

61 )5(5

)1(k

k kk

= 56 +

5

1

62 )5(5

)1(k

k kk

= (-1)0 (5-0)6 +

5

1

6)5(5

)1(k

k kk

50

= Total no. of functions from a 6-element set N to a 5-element set M - Total no. of functions that map nothing to at least one element in M

=

5

0

6)5(5

)1(k

k kk

69


set N to a 5-element set M = {y1, y2, …, y5} are there?Total no. of onto functions from a 6-element set N to a 5-element set M

NM

1

2y1

y2

34

5

6

y3

y4

y5=

5

0

6)5(5

)1(k

k kk

70



NM

1

2y1

y2

34

5

6

y3

y4

y5

=

5

0

6)5(5

)1(k

k kk

71

e.g.23 (Page 37) How many onto functions from a 6-element


NM

1

2y1

y2

34

5

6

y3

y4

y5

n

m

…

n

…

ym

m

=

5

0

6)5(5

)1(k

k kk

n m

nm

mm

1 Inclusion-Exclusion Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong.

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