If (a b 2f a b f if 8b 9 1 a with f a) = b f 8b 9 a f a b ...stacho/macm101-3.pdf · Cartesian (cross) product A B = f(a;b) ja 2A;b 2Bg Relation R from A to B, R A B Function f :

Cartesian (cross) product A× B = {(a, b) | a ∈ A, b ∈ B}Relation R from A to B , R ⊆ A× B

Function f : A→ B , a relation f ⊆ A×B with ∀a ∃!b (a, b) ∈ f

If (a, b) ∈ f we write f (a) = b

One-to-one (injective) function f if ∀b ∃≤1 a with f (a) = b

Onto (surjective) function f if ∀b ∃≥1 a with f (a) = b

Definition 1

For f : A→ B and C ⊆ A we define

f (C ) = {b ∈ B | b = f (a), a ∈ C} = {f (a) | a ∈ C}

The range of f : A→ B is f (A), and f is onto if f (A) = B .

() November 1, 2007 1 / 8

Cartesian (cross) product A× B = {(a, b) | a ∈ A, b ∈ B}Relation R from A to B , R ⊆ A× B

Function f : A→ B , a relation f ⊆ A×B with ∀a ∃!b (a, b) ∈ f

If (a, b) ∈ f we write f (a) = b

One-to-one (injective) function f if ∀b ∃≤1 a with f (a) = b

Onto (surjective) function f if ∀b ∃≥1 a with f (a) = b

Definition 1

For f : A→ B and C ⊆ A we define

f (C ) = {b ∈ B | b = f (a), a ∈ C} = {f (a) | a ∈ C}

The range of f : A→ B is f (A), and f is onto if f (A) = B .

() November 1, 2007 1 / 8

Let A = {1, 2, 3, 4, 5} and B = {w , x , y , z} andf : A→ B be f = {(1, x), (2, x), (3, y), (4, y), (5, z)}.

Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.

a) f (A1) = f ({1}) = {f (1)} = {x}b) f (A2) = f ({1, 2}) = {f (1), f (2)} = {x , x} = {x}c) f (A3) = f ({2, 3}) = {f (2), f (3)} = {x , y}d) f (A4) = f ({2, 4, 5} = {f (2), f (4), f (5)} = {x , y , z}e) f (A3 ∪ A4) = f ({2, 3} ∪ {2, 4, 5}) = f ({2, 3, 4, 5}) ={f (2), f (3), f (4), f (5)} = {x , y , y , z} = {x , y , z}

f) f (A3 ∩ A4) = f ({2, 3} ∩ {2, 4, 5}) = f ({2}) = {f (2)} = {x}g) f (A3) ∪ f (A4) = {x , y} ∪ {x , y , z} = {x , y , z} = f (A3 ∪ A4)

h) f (A3) ∩ f (A4) = {x , y} ∩ {x , y , z} = {x , y} % {x} = f (A3 ∩ A4)

() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8


Let A1 = {1}, A2 = {1, 2}, A3 = {2, 3}, A4 = {2, 4, 5}.




() November 1, 2007 2 / 8

Theorem 2

Let f : A→ B and A1, A2 ⊆ A, then

a) f (A1 ∪ A2) = f (A1) ∪ f (A2)

b) f (A1 ∩ A2) ⊆ f (A1) ∩ f (A2)

c) f (A1 ∩ A2) = f (A1) ∩ f (A2) if f is one-to-one

Proof.a) b ∈ f (A1 ∪ A2) ⇐⇒ b = f (a) for a ∈ A1 ∪ A2 ⇐⇒ b = f (a)

for a ∈ A1 or a ∈ A2 ⇐⇒ b ∈ f (A1) or b ∈ f (A2) ⇐⇒b ∈ f (A1) ∪ f (A2)

b) b ∈ f (A1 ∩ A2) ⇐⇒ b = f (a) for a ∈ A1 ∩ A2 ⇐⇒ b = f (a)

for a ∈ A1 and a ∈ A2/⇐=

=⇒ b ∈ f (A1) and b ∈ f (A2) ⇐⇒b ∈ f (A1) ∩ f (A2)

c) b ∈ f (A1) and b ∈ f (A2) ⇐⇒ b = f (a1) for a1 ∈ A1 and

b = f (a2) for a2 ∈ A2injective⇒a1=a2⇐⇒ b = f (a) for a ∈ A1 ∩ A2

�() November 1, 2007 3 / 8

Theorem 2


a) f (A1 ∪ A2) = f (A1) ∪ f (A2)

b) f (A1 ∩ A2) ⊆ f (A1) ∩ f (A2)






=⇒ b ∈ f (A1) and b ∈ f (A2) ⇐⇒b ∈ f (A1) ∩ f (A2)



�() November 1, 2007 3 / 8

Theorem 2


a) f (A1 ∪ A2) = f (A1) ∪ f (A2)

b) f (A1 ∩ A2) ⊆ f (A1) ∩ f (A2)






=⇒ b ∈ f (A1) and b ∈ f (A2) ⇐⇒b ∈ f (A1) ∩ f (A2)



�() November 1, 2007 3 / 8

Theorem 2


a) f (A1 ∪ A2) = f (A1) ∪ f (A2)

b) f (A1 ∩ A2) ⊆ f (A1) ∩ f (A2)






=⇒ b ∈ f (A1) and b ∈ f (A2) ⇐⇒b ∈ f (A1) ∩ f (A2)



�() November 1, 2007 3 / 8

Definition 3

For f : A→ B and C ⊆ A, let f |C : C → B be the function definedas f |C (a) = f (a) for a ∈ C . We call f |C the restriction of f to C ,and call f an extention of f |C to A.

Let A = {1, 2, 3, 4} and (note: A $ Z $ R)let f : A→ R be defined as f = {(1,−4), (2,−1), (3, 2), (4, 5)},let g : Z→ R be defined as g(x) = 3x − 7 for x ∈ Z andlet h : R→ R be defined as h(x) = 3x − 7 for x ∈ R.

◦◦◦◦

f

••

••

g

��

h a) g is an extention of f (from A) to Zb) f is the restriction of g (from Z) to A

c) h is an extention of f (from A) to Rd) f is the restriction of h (from R) to A

e) h is an extention of g (from Z) to Rf) g is the restriction of h (from R) to Z

() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Definition 3



◦◦◦◦

f

••

••

g

��




() November 1, 2007 4 / 8

Theorem 4

For finite sets A, B with |A| = n and |B | = m, the number of ontofunctions from A to B is

s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n

Proof. By induction on n. For n = 1 easy (note 00 = 1), let n > 1s(n, m) = m · (s(n − 1, m) + s(n − 1, m − 1)) =

= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Theorem 4


s(n, m) =m∑

k=0

(−1)k

(m

k

)(m − k)n


= m(∑m

k=0(−1)k(mk

)(n− k)n−1 +

∑m−1k=0 (−1)k

(m−1

k

)(m− 1− k)n−1

)= m

(∑mk=0(−1)k

(mk

)(n − k)n−1 +

∑mk=1(−1)k−1

(m−1k−1

)(m − k)n−1

)= m ·mn−1 + m

∑mk=1(−1)k

[(mk

)−(m−1k−1

)](m − k)n−1

= mn+∑m

k=1(−1)k(m−1

k

)m

m−k(m−k)n = mn+

∑mk=1(−1)k

(mk

)(m−k)n

=∑m

k=0(−1)k(mk

)(m − k)n �

() November 1, 2007 5 / 8

Exercises.

5.2.3 Let A = {1, 2, 3, 4} and let B = {x , y , z}.a) # of functions from A to B?

b) # of one-to-one functions from A to B

c) # of onto functions from A to B

d) # of functions from B to A?

e) # of one-to-one functions from B to A

f) # of onto functions from B to A

g) # of functions from A to B with f (1) = x

h) # of functions from A to B with f (1) = f (2) = x

i) # of functions from A to B with f (1) = x and f (2) = y

() November 1, 2007 6 / 8

Exercises.

5.2.3 Let A = {1, 2, 3, 4} and let B = {x , y , z}.a) # of functions from A to B? Answer: 34 = 81









() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.


b) # of one-to-one functions from A to B Answer: P(3, 4) = 0








() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.



c) # of onto functions from A to B Answer: 34 − 3 · 24 + 3 = 36







() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.




d) # of functions from B to A? Answer: 43 = 64






() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.





e) # of one-to-one functions from B to A Answer: P(4, 3) = 24





() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.






f) # of onto functions from B to AAnswer: 43 − 4 · 33 + 6 · 23 − 4 = 0




() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.







g) # of functions from A to B with f (1) = x Answer: 33 = 27



() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.








h) # of functions from A to B with f (1) = f (2) = xAnswer: 32 = 9


() November 1, 2007 6 / 8

Exercises.










() November 1, 2007 6 / 8

Exercises.









i) # of functions from A to B with f (1) = x and f (2) = yAnswer: 32 = 9

() November 1, 2007 6 / 8

Theorem 5

The number of ways of distributing n distinct objects into mcontainers where no container is empty is S(n, m) = 1

m!· s(n, m)

Proof. Each distribution defines a surjective mapping from{1, . . . , n} to {1, . . . m}, and there are m! different ways ofnumbering the containers. �

Example. 5.3.8 Suppose we have seven differently coloured balls andfour containers I , II , III , IV :

a) # ways to distribute balls

b) # ways to distribute balls so that no container is empty

c) # ways to distribute balls so that no container is empty and theblue ball is in II

d) a) and b) if we remove labels from the containers

() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)



a) # ways to distribute balls




() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)



a) # ways to distribute balls Answer: 47 = 16384




() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)







() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)




b) # ways to distribute balls so that no container is emptyAnswer: s(7, 4) = 4! · 350 = 8400



() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)







() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)





c) # ways to distribute balls so that no container is empty and theblue ball is in II Answer: s(6, 4) = 4! · 65 = 1560


() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)







() November 1, 2007 7 / 8

Theorem 5


m!· s(n, m)






d) a) and b) if we remove labels from the containersAnswer: a) S(7, 1) + S(7, 2) + S(7, 3) + S(7, 4) = 715; b) 350

() November 1, 2007 7 / 8

Exercises5.3.13 How many factorisations of 156,009 into

a) two factors greater than one

b) two or more factors greater than one

c) a) and b) for any n = p1 · . . . · pk where all pi are different

5.3.16 Ten candidates C1, . . . , C10 for senior class president

a) How many possible outcomes of the election (no ties)?

b) How many possible outcomes with possible ties?

c) b) where exactly three candidates tie for first place?

d) a), b) and c) where C3 in first place?

() November 1, 2007 8 / 8

Exercises5.3.13 How many factorisations of 156,009 into

a) two factors greater than one








() November 1, 2007 8 / 8

Exercises5.3.13 How many factorisations of 156,009 = 3 · 7 · 17 · 19 · 23 into

a) two factors greater than one Answer: S(5, 2) = 15








() November 1, 2007 8 / 8










() November 1, 2007 8 / 8



b) two or more factors greater than oneAnswer: S(5, 2) + S(5, 3) + S(5, 4) + S(5, 5) = 51







() November 1, 2007 8 / 8










() November 1, 2007 8 / 8




c) a) and b) for any n = p1 · . . . · pk where all pi are differenta) S(k , 2) b) S(k , 2) + S(k , 3) + . . . + S(k , k) =

∑ki=2 S(k , i)






() November 1, 2007 8 / 8





∑ki=2 S(k , i)






() November 1, 2007 8 / 8





∑ki=2 S(k , i)






() November 1, 2007 8 / 8





∑ki=2 S(k , i)


a) How many possible outcomes of the election (no ties)? 10!




() November 1, 2007 8 / 8





∑ki=2 S(k , i)






() November 1, 2007 8 / 8





∑ki=2 S(k , i)



b) How many possible outcomes with possible ties?Answer: s(10, 1) + s(10, 2) + . . . + s(10, 10) = 102, 247, 563



() November 1, 2007 8 / 8





∑ki=2 S(k , i)






() November 1, 2007 8 / 8





∑ki=2 S(k , i)




c) b) where exactly three candidates tie for first place?Answer:

(103

)∑7i=1 s(7, i) = 120 · 47, 293 = 5, 675, 160


() November 1, 2007 8 / 8





∑ki=2 S(k , i)





(103

)∑7i=1 s(7, i) = 120 · 47, 293 = 5, 675, 160


() November 1, 2007 8 / 8





∑ki=2 S(k , i)





(103

)∑7i=1 s(7, i) = 120 · 47, 293 = 5, 675, 160

d) a), b) and c) where C3 in first place?Answer: a) 9! b)

∑9i=1 s(9, i) c)

(92

)∑7i=1 s(7, i)

() November 1, 2007 8 / 8

Documents

If (a b 2f a b f if 8b 9 1 a with f a) = b f 8b 9 a f a b ...stacho/macm101-3.pdf · Cartesian (cross) product A B = f(a;b) ja 2A;b 2Bg Relation R from A to B, R A B Function f :