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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
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)?
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
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)?
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
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)?
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
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)?
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)
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 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
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 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)
We further re-write as
)()()( 321
2
1
3
1
3
1
3
1
EEEPEEPEPEPi ij
jii
ii
i
We further re-write as
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: 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.
Step 2(b): According to P(n-1), we deduce that P(n) is true.
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
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: 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.
Step 2(b): According to P(n-1), we deduce that P(n) is true.
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
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(
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
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(
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
Objective:
Consider
n
iiEP
1
n
n
iin
n
ii EEPEPEP )(
1
1
1
1
Inductive Hypothesis:
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(
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
Objective:
Consider
n
iiEP
1
Inductive Hypothesis:
Let Gi = Ei En for i < n
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(
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
Objective:
Consider
n
iiEP
1
Inductive Hypothesis:
Let Gi = Ei En for i < n
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(
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
Objective:
Consider
n
iiEP
1
Inductive Hypothesis:
Let Gi = Ei En for i < n
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
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
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
21)...()1(Objective:
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
21)...()1(Objective:
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
21)...()1(Objective:
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
21)...()1(Objective:
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
(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?
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
(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?
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)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)
= (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!
(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?
(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?
(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?
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
(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?
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)
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
(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?
1
n(n-k)!
n!
P(at least one person gets his own backpack)
Let Ei be the event that student i gets his own backpack back.
!
1)1(
5
1
1
kk
k
47
e.g.12
(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?
1
n(n-k)!
n!
P(at least one person gets his own backpack)
Let Ei be the event that student i gets his own backpack back.
!
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?
(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?
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?
(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?
1
n(n-k)!
n!
niiiiii
iii
n
k
k
k
k
kEEEP
...1:,...,,1
1
21
21
21)...()1(
P(at least one person gets his own backpack)
= P(E1 U E2 U … U En)
50
e.g.14
(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?
1
n(n-k)!
n!
niiiiii
iii
n
k
k
k
k
kEEEP
...1:,...,,1
1
21
21
21)...()1(
P(at least one person gets his own backpack)
= 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)
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 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
(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?
1
n(n-k)!
n!
P(at least one person gets his own backpack)
Let Ei be the event that student i gets his own backpack back.
!
1)1(
1
1
k
n
k
k
52
e.g.12
(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?
1
n(n-k)!
n!
P(at least one person gets his own backpack)
Let Ei be the event that student i gets his own backpack back.
!
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?
(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?
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)!
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 – P(at least one person 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
e.g.15P(nobody gets his own backpack)
1e if n is a large number
!
)1(...
!3
)1(
!2
)1()1(1
32
n
n
56
e.g.15P(nobody gets his own backpack)
1e if n is a large number
!
)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
e.g.18 (Page 33) How many functions from a 6-
element set N to a 5-element set M = {y1, y2, …, y5} map nothing to y1?
NM
1
2y1
y2
4 choices
4 choices
Total no. of functions = 4 x 4 x 4 x 4 x 4 x 4= 46
34 choices
4
5
6
4 choices
4 choices
4 choices
y3
y4
y5
60
e.g.19 (Page 33) How many functions from a 6-
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
Total no. of functions = 3 x 3 x 3 x 3 x 3 x 3= 36
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?
Total no. of functions that map nothing to a given set K of k elements in M= (5-k)6
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
set N to a 5-element set M = {y1, y2, …, y5} map nothing to at least one element in M?
Total no. of functions that map nothing to a given set K of k elements in M= (5-k)6
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
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
set N to a 5-element set M = {y1, y2, …, y5} map nothing to at least one element in M?
Total no. of functions that map nothing to a given set K of k elements in M= (5-k)6
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
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?
Total no. of functions that map nothing to at least one element in M=
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
Total no. of functions that map nothing to at least one element in M=
5
1
61 )5(5
)1(k
k kk
67
e.g.22
Not onto function (or not surjection)N M
NS M
Total no. of functions that map nothing to at least one element in 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?
Total no. of functions that map nothing to at least one element in M=
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
e.g.22 How many onto functions from a 6-element
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
e.g.22 How many onto functions from a 6-element
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
71
e.g.23 (Page 37) How many onto functions from a 6-element
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
n
m
…
n
…
ym
m
=
5
0
6)5(5
)1(k
k kk
n m
nm
mm
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