Power set
2.7 Stirling Number of the Second Kind { n
k
}
The Stirling number of the second kind denoted by {n
k
} is the number ofunordered partitionsof then-setA={1,2, . . . , n}into
kmutually disjoint non-empty subsets. Note that the empty set is notpermitted in the partition. (In the partitions corresponding to a multinomial coefficient, empty sets are allowed.) Such a partition is called a k-partition of the set A.
The curly braces in {n
k
} (read n subset k) can be easily re- membered, since curly braces denote sets. The following example illustrates the Stirling number of the second kind.
Example 2.7.1 (Stirling number of the second kind): What is the value of{4
2
}?
Solution: Consider the 4-set A = {1,2,3,4}. The pos- sible partitions of the set A into two non-empty subsets are: {1,2,3} {4}; {1,2,4} {3}; {1,3,4} {2} ; {2,3,4} {1};
{1,2} {3,4}; {1,3} {2,4}; {1,4} {2,3}. There are 7 partitions of A into two non-empty subsets. Therefore by definition, {4
2
}= 7.
{n 1
}
= 1 for n≥1
because the number of unordered partitions of ann-setAintoone mutually disjoint non-empty subset is 1(the set {A} is the only partition of A). Again
{n n
}
= 1 for n≥1
since the only unordered partition of then-set A= {1,2, . . . , n} inton mutually disjoint subsets is the partition {A1, A2, . . . , An} whereAi ={i} for all i= 1,2, . . . , n.
We define {0
0
} = 1. Note that{n
k
} = 0 if k > n.
The initial and boundary values of the binomial coefficients, Stirling numbers of the first and second kind may be written suc- cinctly with the help of theKronecker delta δij where
δij =
{ 1 if i=j 0 otherwise Kronecker’s famous citation:
2.7. STIRLING NUMBER OF THE SECOND KIND {N
K
} 115
God created the natural numbers, all the rest is the work of man.
Kronecker Another interesting quotation:
Man created the integers, all else is Dieudonn´e.
R. K. Guy Using the notation δij, we write for non-negative integer n,
(0 n
)
= [0
n ]
= {0
n }
=δn0. Furthermore,
(n n
)
= [n
n ]
= {n
n }
= 1 =δnn.
We shall now prove the following addition formula (recurrence formula) involving Stirling numbers of the second kind.
Lemma 2.7.1 (Recurrence formula for Stirling number of the sec- ond kind):
For non-negative integers n and k we have:
{n+ 1 k
}
=k {n
k }
+ { n
k−1 }
Proof. Consider the (n+ 1)-set A = {1,2, . . . , n+ 1}. By defi- nition, {n+1
k
} is the number of unordered partitions of the set A into k mutually disjoint non-empty subsets of A. We distinguish two types ofk-partitions of the (n+ 1)-setA.
Case 1: k-Partitions ofAhaving the singleton set{n+ 1}as a member of the partition. (This set can be any other singleton set.
If 1,2, . . . , n, n+ 1 are professors in the Mathematics Department, then n+ 1 can be considered as the “Head” of the Mathematics Department.)
We are interested in counting the number of such partitions in terms of n. By definition, { n
k−1
} is the number of unordered
(k−1)-partitions of the n-set A\ {n+ 1}. Each (k−1)-partition {A1, A2, . . . , Ak−1}of then-setA\{n+1}defines thek-partition {A1, A2, . . . , Ak−1} ∪ { {n+ 1} }of the (n+ 1)-setA.Conversely, eachk-partition of the (n+ 1)-setA having the singleton set{n+ 1}as a member, defines a (k−1)-partition (obtained by removing the set {n+ 1} from the k-partition) of the n-set A\ {n+ 1}. Hence byProposition 2.2.1, the number ofk-partitions ofAhaving the singleton set{n+ 1}as a member of the partition is the same as the number of (k−1)-partitions of then-set A\ {n+ 1}which is by definition { n
k−1
}
Case 2: k-partitions of A having no singleton subset {n+ 1} as a member, that is, partitions in which subsets containing the element n+ 1 must have at least two elements.
Let us count the number of such partitions in terms ofn.
Consider a k-partition {A1, A2, . . . , Ak} of A such that the subsetAi containing the elementn+1 and the number of elements of Ai is ≥2. Now the {A1, A2, . . . , Ai \ {n+ 1}, . . . , Ak} is a k- partition of then-set A\ {n+ 1}, sinceAi\ {n+ 1} ̸=∅.
Conversely, consider a k-partition {A1, A2, . . . , Ak} of the n- setA\{n+ 1}.This partition definesk k-partitions{A1, A2, . . . , Ai∪ {n+ 1}, . . . , Ak} of the set (n+ 1)-setA for i= 1,2, . . . , k.
Therefore the number of desired partitions in this case isk{n
k
}. Since case 1 and case 2 are mutually disjoint, we have by the rule of sum (see Fact 2.2.0.1),
{n+ 1 k
}
=k {n
k }
+ { n
k−1 }
Thus the proof is complete.
The above recurrence relation allows us to compute the num- bers{n
k
} starting from the initial conditions {0
0
}= 0 and {n
1
}= 1 forn ≥1. The following table, called the Stirling triangle of the second kind, constructs the numbers {n
k
} iteratively. To obtain thek-th number in the (n+ 1)-th line, (assuming that nlines have already been constructed), we multiply by k the kth number of the nth line and add it to the (k−1)-th number of the n-th line.
The table may be continued to any number of lines. Note that the first line is the 0-th line.
2.7. STIRLING NUMBER OF THE SECOND KIND {N
K
} 117
For example, the 4-th line gives the numbers {4
1
}; {4
2
}; {4
3
}; {4
4
}; and {4
5
}.
Table 2.7: Stirling triangle of the second kind 1
1 0
1 1 0
1 3 1 0
1 7 6 1 0
...
Example 2.7.2 (Stirling numbers of the first and second kinds): Prove that{n
k
} ≤[n
k
].
Combinatorial proof: Every partition of [n] into nonempty sub- sets induces at least one cycle. For example, the singleton sets{1} and {1,2} give only one cycle (1) and (1,2) = (2,1) respectively.
All other sets of at least 3 elements give more than one cycle.
For example, the set of three elements{1,2,3} give two different cycles (1,2,3) and (1,3,2).Hence the claimed inequality.
Example 2.7.3 (Stirling numbers of the first and second kinds):
Prove that [
n n−1
]
= { n
n−1 }
= (n
2 )
Solution: By definition, [ n
n−1
] is the coefficient of xn−1 in the product x(x+ 1)(x+ 2)· · ·(x+n−1), that is, the coefficient of xn−2 in
(x+ 1)(x+ 2)(x+ 3)· · ·(x+n−1) =xn−2 +(1 + 2 + 3 +· · ·+n−1)xn−2+· · ·+ (n−1)!
Hence [ n
n−1
]= 1 + 2 +· · ·+ (n−1) = (n−1)n/2 =(n
2
). Now, by definition, { n
n−1
} is the number of unordered parti- tions of then-setA ={1,2, . . . , n} into (n−1) mutually disjoint
non-empty subsets of A. The sets of such a partition are of the form {A1, A2, . . . , Ai, . . . , An−1} where exactly one of the Ai’s is a two-element subset of A and all others are singleton subsets of A.Note that once we pick a 2-subset Ai ofA, all other sets in the partition are uniquely determined.
Therefore, the number of such partitions is the same as the number of two-element subsets of then-set, which by definition is (n
2
).
Example 2.7.4 (Property):
Let p be a prime number. Then prove that {p
k
} for all k with 1< k < p.
Solution: Similar to Example 2.6.13.
Example 2.7.5 (Stirling number of the second kind): Prove that for non-negative integern,
{n+ 1 2
}
= 2n−1.
Solution: By definition, {n+1
2
} is the number of unordered parti- tions of the (n+ 1)-set A = {1,2, . . . , n+ 1} into two mutually disjoint non-empty subsets of A.
First we count the number of ordered 2-partitions of the set A (empty set allowed as a part of the partition.)
By definition of the multinomial coefficient, the number of ordered (k, n − k) 2-partition of the n-set A ( 0 ≤ k ≤ n) is ( n
k,n−k
)= k!(nn!−k)! =(n
k
). Hence the number of ordered 2-partitions of the (n+ 1)-set is
∑n k=1
(n k
)
= 2n+1 by Corollary 2.4.1 .
These 2n+1 ordered 2-partitions of A include the special two or- dered partitions (∅, A) and (A,∅) of which one part is empty. If we remove these two partitions, then we have the number of ordered 2-partitions (A1, A2) (withA1 ̸=∅ and A2 ̸=∅) of the (n+ 1)-set which is 2n+1−2.
2.7. STIRLING NUMBER OF THE SECOND KIND {N
K
} 119
But the two ordered 2-partitions (A1, A2) and (A2, A1) define only one unorderedpartition {A1, A2}. Therefore, the number of unordered partitions of the set A into two mutually disjoint non- empty subsets is 12 ×(2n+1−2) = 2n−1.
Number of surjective functions from an n-set onto an m-set (n ≥ m): We shall now find the number of surjective functions from ann-set onto anm-set using the Stirling number of the second kind. The following example illustrates how a surjective functionf from ann-setX onto anm-set Y induces an unordered partition of the domain set X into mutually disjoint non-empty subsets of X.
First we recall the idea of an inverse image of any function f :X →Y.For an element y∈Y,f−1(y) is the set of all elements x∈X whose image under f is y. Symbolically,
f−1(y) = {x∈X |f(x) = y}
Example 2.7.6 (Partition induced by a surjective function): LetX ={p1, p2, . . . p10} be a set of professors in the Department of Mathematics and let Y = {r1, r2, . . . , r6} be the rooms in the Department of Mathematics in a university. Consider the surjec- tive functionf :X →Y which assigns to each professor his room in the Department of Mathematics. f is written in the matrix form where
f =
( p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 r3 r6 r3 r1 r2 r4 r5 r4 r3 r6
)
The function is clearly surjective because every room is occupied by at least one professor. Now for roomr∈Y,f−1(r) is the set of professors occupying the roomrand this set is non-empty because f is surjective. The partition induced by the function f is
f−1(r1)∪f−1(r2)∪· · ·∪f(−1)(r6) = {p4}∪{p5}∪{p1, p3, p9}∪
{p6, p8}{p7} ∪ {p2, p10}.
With this example in mind, we state the following proposition.
Proposition 2.7.1:
Letf :X →Y be a surjective function from an n-set X onto an m-set Y.Thenf induces an unordered partition of the domain set X into m mutually disjoint non-empty subsets of X.
Proof. IfY ={y1, y2, . . . , ym}, then the required partition of the setX intom mutually disjoint non-empty subsets is
{f−1(y1), f−1(y2), . . . , f−1(ym)}
The following theorem gives the number of surjective functions from ann-set onto an m-set.
Theorem 2.7.1:
The number of surjective functions from an n-set onto an m-set (n ≥m) is {n
m
}m!.
Proof. Let the domain set be then-set X ={x1, x2, . . . , xn}and let the range set be the m-set Y = {y1, y2, . . . , ym}. By Propo- sition 2.7.1, every surjective function from X onto Y induces a partition of X into mutually disjoint non-empty subsets ofX.
Now consider an unordered m-partition {X1, X2, . . . , Xm} of the setX.Consideranypermutation (written in one-line notation) pi = (yi1yi2. . . yim) of the m-set Y. This permutation defines a surjective functionpi fromX onto Y where
pi(xj) =yij if xj ∈Xj, 1 ≤j ≤n
That is, pi maps every element of the subsetXj onto the element yij.Since there arem! permutations possible on them-setY, each m-partition{X1, X2, . . . , Xm}of the setX inducesm! surjections fromX ontoY.But the number partitions possible of then setX intom mutually disjoint non-empty sets is the Stirling number of the second kind{n
m
}, by definition. Therefore by the product rule (see Proposition 2.2.2), the number of surjections from the n-set X onto the m-set Y is{n
m
}m!.
2.7. STIRLING NUMBER OF THE SECOND KIND {N
K
} 121
Example 2.7.7:
The number of surjective functions from the 4-set X = {x1, x2, x3, x4} onto the 3-set Y = {y1, y2, y3} is {4
3
}3!.
From the Table 2.7, {4
3
} = 6. Hence the number of surjections is 6×3! = 36.
Example 2.7.8:
Find the number of ways to partition n elements in m different boxesB1, B2, . . . , Bm in such a way that k of these boxes are non- empty and m−k of these boxes are empty.
Solution:
The number of ways to pick k boxes out of m boxes is (m
k
) ways. The number of ways to partition (unordered partition) n elements into k mutually disjoint subsets is the Stirling number {n
k
}. Hence, the number of ways to put n elements into k boxes Bi1, Bi2, . . . , Bik such that no box is empty is the same as number oforderedpartitions ofn-elements intoknon-empty subsets which isk!{n
k
}(since each unordered partition intoknon-empty subsets gives rise tok! ordered partitions). Therefore by the product rule of Proposition 2.2.2, the desired number is k!{n
k
} (m
k
).
Example 2.7.9:
Find the number of equivalence relations that can be defined on the set [n] ={1,2, . . . , n}.
Solution: We know that every equivalence relation on [n] parti- tions the set [n] into nonempty mutually disjoint subsets called the equivalence classes and given any partition of [n] into nonempty mutually disjoint subsets, we can define an equivalence relation on [n] whose equivalence classes are exactly the given partitions of the set.
Hence by the definition of the Stirling number of the second kind, the number of equivalence relations on [n] is the number of possible partitions of [n] into k (1 ≤ k ≤ n) nonempty mutually disjoint subsets. This number is
∑n k=1
{n k
} .
Corollary 2.7.1.1:
For positive integersm and n we have the equality mn=
∑n k=1
{n k
} mk
Proof. ByTheorem 2.2.2, the left-hand side of the equality mn is the number of functions from an n-set X = {x1, x2, . . . , xn} to the m-set Y ={y1, y2, . . . , ym}.
Now let us compute the number of functions from X to Y in a different way. Let us first observe that any function f :X →Y becomes a surjective function if we restrict the co-domain set Y to the subsetf(X) ={f(x)|x∈X}, the set of all images of the elements of the domain set X.Furthermore, 1 ≤ |f(X)| ≤n.
Consider a k-subset T of Y where k is an integer such that 1 ≤ k ≤ n. By Theorem 2.7.1, the number of surjections from X onto the k-set T is {n
k
}k!. But a k-subset T of the m-set Y can be chosen in(m
k
), by the definition of the binomial coefficient.
Therefore, by the product rule (Proposition 2.2.2), the number of surjections from X onto a k-subset of Y is {n
k
}k!(m
k
). Since the integer k may assume any value between 1 and n, we have that the number of surjections from X onto a subset of Y is
∑n k=1
{n k
} k!
(m k
)
=
∑n k=1
{n k
} mk
Example 2.7.10:
Verify the equality in theCorollary 2.7.1.1, for n= 3 and m= 4.
The left-hand side is mn= 43 = 64.
The right-hand side of the equality is
∑n k=1
{n k
}
mk =
∑3 k=1
{3 k
} 4k
= {3
1 }
4 + {3
2 }
(4)(3) + {3
3 }
(4)(3)(2)
2.7. STIRLING NUMBER OF THE SECOND KIND {N
K
} 123
= (1)(4) + (3)(4)(3) + (1)(4)(3)(2)
= 4 + 36 + 24
= 64
Consider the equality mn = ∑n k=1
{n
k
}mk of the Corollary 2.7.1.1. This equation can be proved to be true for any real num- berx. To do this, we need the following simple lemma concerning polynomials. To prove the lemma, we need the Fundamental The- orem of Algebra, which we state without proof.
Theorem 2.7.2 (Fundamental theorem of algebra):
Consider a polynomialp(z) =p0+p1z+p2z2+· · ·+pnznof degree n with complex coefficients pi. Then p(z) has exactly n complex roots (distinct or coincident).
The following lemma is very useful to extend the validity of some equations from integers to all real numbers.
Lemma 2.7.2:
Letp(x) =p0+p1x+· · ·+pnxn and q(x) = q0+q1x+· · ·+qnxn be two non-zero polynomials of degree n. If p(x) =q(x) for n+ 1 distinct real numbersr0, r1, . . . , rn then the polynomialsp(x) and q(x) are identically equal, symbolically, p(x)≡q(x).
Proof. p(x) =q(x) for x=r0, r1, . . . , rn. Therefore r(x) =p(x)− q(x) is a polynomial of degree at mostn. Moreover, the polynomial r(x) of degree at most n is such that s(ri) = p(ri)−q(ri) = 0 for i = 0,1, . . . , n. In other words, the polynomial s(x) which is of degree at most n has n + 1 distinct roots. Hence by the fundamental theorem of algebra 2.7.2, the polynomial s(x) must be identically zero, that is, s(x) = 0 for all real numbers. That is, p(x)−q(x)≡0, that is, p(x)≡q(x).
Corollary 2.7.2.1:
If x is any real number and n is a positive integer then we have
the equality
xn=
∑n k=1
{n k
} xk
Proof. By Theorem 2.7.1.1, we have the equality mn =
∑n k=1
{n k
}
mk for positive integers m, n.
The left-hand side and the right-hand side of the above equation may be viewed as polynomials of degreenin the variablem.These two polynomials of degree n in the variable m assume the same values forn+ 1 distinct values ofmwherem= 0,1,2, . . . , n(since for m = 0, both polynomials are 0.) Hence by Lemma 2.7.2, the left-hand side isidentically equal to the right-hand side. This proves the corollary.
Theorem 2.7.3:
{n+ 1 k
}
=
∑n p=0
(n p
) { p k−1
} .
Proof. Consider the (n + 1)-set A = {1,2, . . . , n, n+ 1} and a partitionP ={P1, P2, . . . , Pk} ofA intok mutually disjoint non- empty subsets.
LetPibe the set containing the integern+1.Then byremoving the setPi fromP, we get the partition{P1, P2, . . . , Pi−1, Pi+1, Pk} of the subset A\Pi into k−1 mutually disjoint subsets.
Conversely, given a subsetS ofA withp elements (0≤p≤n) not containing the integern+ 1 and a partitionP ={P1, P2, . . . , Pk−1}ofS intok−1 mutually disjoint subsets, we get a partition P ∪(A\S) of A byadding the set A\S toP.
But a subsetS ofAconsisting ofpelements not containing the integern+ 1 may be picked in (n
p
) ways. Thus we have established a one-to-one correspondence between the family of all partitions of the setAand the family of all partitions of the subsets ofAnot containing the elementn+ 1.