Discrete Mathematics

Discrete Mathematics

WEN-CHING LIEN Department of Mathematics National Cheng Kung University

2008

5.3: Onto Functions: Stirling Numbers of the Second Kind

Definition (5.9)

A function f :A→B, is called onto, or surjective, if f(A) =B that is, if for all b ∈B there is at least one a∈A with f(a) =b.

Example (5.22)

If A= {x,y,z}and B= {1,2}, then all functions f :A→B, are onto except

f₁= {(x,1), (y,1), (z,1)},and f₂= {(x,2), (y,2), (z,2)},the constant functions.

So there are|B|^|A|−2=2³−2=6 onto functions from A to B.

5.3: Onto Functions: Stirling Numbers of the Second Kind

Definition (5.9)

A function f :A→B, is called onto, or surjective, if f(A) =B that is, if for all b ∈B there is at least one a∈A with f(a) =b.

Example (5.22)

If A= {x,y,z}and B= {1,2}, then all functions f :A→B, are onto except

f₁= {(x,1), (y,1), (z,1)},and f₂= {(x,2), (y,2), (z,2)},the constant functions.

So there are|B|^|A|−2=2³−2=6 onto functions from A to B.

5.3: Onto Functions: Stirling Numbers of the Second Kind

Definition (5.9)

A function f :A→B, is called onto, or surjective, if f(A) =B that is, if for all b ∈B there is at least one a∈A with f(a) =b.

Example (5.22)

If A= {x,y,z}and B= {1,2}, then all functions f :A→B, are onto except

f₁= {(x,1), (y,1), (z,1)},and f₂= {(x,2), (y,2), (z,2)},the constant functions.

So there are|B|^|A|−2=2³−2=6 onto functions from A to B.

5.3: Onto Functions: Stirling Numbers of the Second Kind

Definition (5.9)

A function f :A→B, is called onto, or surjective, if f(A) =B that is, if for all b ∈B there is at least one a∈A with f(a) =b.

Example (5.22)

If A= {x,y,z}and B= {1,2}, then all functions f :A→B, are onto except

f₁= {(x,1), (y,1), (z,1)},and f₂= {(x,2), (y,2), (z,2)},the constant functions.

So there are|B|^|A|−2=2³−2=6 onto functions from A to B.

5.3: Onto Functions: Stirling Numbers of the Second Kind

Definition (5.9)

A function f :A→B, is called onto, or surjective, if f(A) =B that is, if for all b ∈B there is at least one a∈A with f(a) =b.

Example (5.22)

If A= {x,y,z}and B= {1,2}, then all functions f :A→B, are onto except

f₁= {(x,1), (y,1), (z,1)},and f₂= {(x,2), (y,2), (z,2)},the constant functions.

So there are|B|^|A|−2=2³−2=6 onto functions from A to B.

Numbers of onto function from A to B:

For any sets A, B with |A| =m and|B| =n,there are

n n

n^m− _n−1ⁿ
(n−1)^m+ _n−2ⁿ
(n−2)^m− · · · +(−1)^{n−2 n}₂

2^m+ (−1)^{n−1 n}₁

1^m=Pn−1

k=0(−1)^n−k _n−kⁿ
(n−k)^m

=

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m onto functions from A to B.

Numbers of onto function from A to B:

For any sets A, B with |A| =m and|B| =n,there are

n n

n^m− _n−1ⁿ
(n−1)^m+ _n−2ⁿ
(n−2)^m− · · · +(−1)^{n−2 n}₂

2^m+ (−1)^{n−1 n}₁

1^m=Pn−1

k=0(−1)^n−k _n−kⁿ
(n−k)^m

=

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m

onto functions from A to B.

Numbers of onto function from A to B:

For any sets A, B with |A| =m and|B| =n,there are

n n

n^m− _n−1ⁿ
(n−1)^m+ _n−2ⁿ
(n−2)^m− · · · +(−1)^{n−2 n}₂

2^m+ (−1)^{n−1 n}₁

1^m=Pn−1

k=0(−1)^n−k _n−kⁿ
(n−k)^m

=

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m

onto functions from A to B.

Numbers of onto function from A to B:

For any sets A, B with |A| =m and|B| =n,there are

n n

n^m− _n−1ⁿ
(n−1)^m+ _n−2ⁿ
(n−2)^m− · · · +(−1)^{n−2 n}₂

2^m+ (−1)^{n−1 n}₁

1^m=Pn−1

k=0(−1)^n−k _n−kⁿ
(n−k)^m

=

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m onto functions from A to B.

Rule:

For m ≥n there arePn

k=0(−1)^n−k _n−kⁿ
(n−k)^m ways to distribute m distinct objects into n numbered(but otherwise identical) containers with no container left empty.

Removing the numbers on the containers, so that they are now identical in appearance, we find that one distribution into these n (nonempty) identical containers corresponds with n!such distributions into the numbered containers.

...continued

Rule:

For m ≥n there arePn

k=0(−1)^n−k _n−kⁿ
(n−k)^m ways to distribute m distinct objects into n numbered(but otherwise identical) containers with no container left empty.

Removing the numbers on the containers, so that they are now identical in appearance, we find that one distribution into these n (nonempty) identical containers corresponds with n!such distributions into the numbered containers.

...continued

So the number of ways in which it is possible to distribute the m distinct objects into n identical containers, with no container left empty, is

1 n!

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m

This will be denoted by S(m,n)and is called a Stirling number of the second kind.

We note that for |A| =m≥n= |B|,there are n! ·S(m,n)onto functions from A to B.

So the number of ways in which it is possible to distribute the m distinct objects into n identical containers, with no container left empty, is

1 n!

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m

This will be denoted by S(m,n)and is called a Stirling number of the second kind.

We note that for |A| =m≥n= |B|,there are n! ·S(m,n)onto functions from A to B.

So the number of ways in which it is possible to distribute the m distinct objects into n identical containers, with no container left empty, is

1 n!

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m

This will be denoted by S(m,n)and is called a Stirling number of the second kind.

We note that for |A| =m≥n= |B|,there are n! ·S(m,n)onto functions from A to B.

So the number of ways in which it is possible to distribute the m distinct objects into n identical containers, with no container left empty, is

1 n!

n

X

k=0

(−1)^n−k

 n n−k



(n−k)^m

This will be denoted by S(m,n)and is called a Stirling number of the second kind.

We note that for |A| =m≥n= |B|,there are n! ·S(m,n)onto functions from A to B.

Example (5.27) For m ≥n,Pn

i=1S(m,i)is the number of possible ways to distribute m distinct objects into n identical containers with empty containers allowed.From the fourth row of Table 5.1

we see that there are 1+7+6=14 ways to distribute the

we see that there are 1+7+6=14 ways to distribute the

Theorem (5.3)

Let m, n be positive integers with 1<n≤m.Then S(m+1,n) =S(m,n−1) +nS(m,n).

Theorem (5.3)

Let m, n be positive integers with 1<n≤m.Then S(m+1,n) =S(m,n−1) +nS(m,n).

Proof:

Let A= {a1,a₂, ...,a_m,a_m+1}Then S(m+1,n)counts the number of ways in which the objects of A can be

distributed among n identical containers, with no container left empty.

1 a_m+1is in a container by itself.

(i) There are S(m,n−1)ways of distributing a1,a2, ...,am

objects among n−1 identical containers.

(ii) Placing am+1in the remaining empty (nth) container results in S(m,n−1)ways counted in S(m+1,n).

2 a_m+1is in the same container as another object.

Proof:

Let A= {a1,a₂, ...,a_m,a_m+1}Then S(m+1,n)counts the number of ways in which the objects of A can be

distributed among n identical containers, with no container left empty.

1 a_m+1is in a container by itself.

(i) There are S(m,n−1)ways of distributing a1,a2, ...,am

objects among n−1 identical containers.

(ii) Placing am+1in the remaining empty (nth) container results in S(m,n−1)ways counted in S(m+1,n).

2 a_m+1is in the same container as another object.

Proof:

Let A= {a1,a₂, ...,a_m,a_m+1}Then S(m+1,n)counts the number of ways in which the objects of A can be

distributed among n identical containers, with no container left empty.

1 a_m+1is in a container by itself.

(i) There are S(m,n−1)ways of distributing a1,a2, ...,am

objects among n−1 identical containers.

(ii) Placing am+1in the remaining empty (nth) container results in S(m,n−1)ways counted in S(m+1,n).

2 a_m+1is in the same container as another object.

Proof:

Let A= {a1,a₂, ...,a_m,a_m+1}Then S(m+1,n)counts the number of ways in which the objects of A can be

distributed among n identical containers, with no container left empty.

1 a_m+1is in a container by itself.

(i) There are S(m,n−1)ways of distributing a1,a2, ...,am

objects among n−1 identical containers.

(ii) Placing am+1in the remaining empty (nth) container results in S(m,n−1)ways counted in S(m+1,n).

2 a_m+1is in the same container as another object.

Proof:

Let A= {a1,a₂, ...,a_m,a_m+1}Then S(m+1,n)counts the number of ways in which the objects of A can be

distributed among n identical containers, with no container left empty.

1 a_m+1is in a container by itself.

(i) There are S(m,n−1)ways of distributing a1,a2, ...,am

objects among n−1 identical containers.

(ii) Placing am+1in the remaining empty (nth) container results in S(m,n−1)ways counted in S(m+1,n).

2 a_m+1is in the same container as another object.

(i) There are S(m,n)ways of distributing a₁,a₂, ...,amobjects among n identical containers.

(ii) Placing a_m+1in the n distinct container results in nS(m,n) ways counted in S(m+1,n).

∴Totally,S(m+1,n) =S(m,n−1) +nS(m,n).

(i) There are S(m,n)ways of distributing a₁,a₂, ...,amobjects among n identical containers.

(ii) Placing a_m+1in the n distinct container results in nS(m,n) ways counted in S(m+1,n).

∴Totally,S(m+1,n) =S(m,n−1) +nS(m,n).

(i) There are S(m,n)ways of distributing a₁,a₂, ...,amobjects among n identical containers.

(ii) Placing a_m+1in the n distinct container results in nS(m,n) ways counted in S(m+1,n).

∴Totally,S(m+1,n) =S(m,n−1) +nS(m,n).

Thank you.