MathDB
Problems
Contests
Undergraduate contests
Miklós Schweitzer
1975 Miklós Schweitzer
2
2
Part of
1975 Miklós Schweitzer
Problems
(1)
Miklos Schweitzer 1975_2
Source:
12/30/2008
Let
A
n
\mathcal{A}_n
A
n
denote the set of all mappings
f
:
{
1
,
2
,
…
,
n
}
→
{
1
,
2
,
…
,
n
}
f: \{1,2,\ldots ,n \} \rightarrow \{1,2,\ldots, n \}
f
:
{
1
,
2
,
…
,
n
}
→
{
1
,
2
,
…
,
n
}
such that
f
−
1
(
i
)
:
=
{
k
:
f
(
k
)
=
i
}
≠
∅
f^{-1}(i) :=\{ k \colon f(k)=i\ \} \neq \varnothing
f
−
1
(
i
)
:=
{
k
:
f
(
k
)
=
i
}
=
∅
implies
f
−
1
(
j
)
≠
∅
,
j
∈
{
1
,
2
,
…
,
i
}
.
f^{-1}(j) \neq \varnothing, j \in \{1,2,\ldots, i \} .
f
−
1
(
j
)
=
∅
,
j
∈
{
1
,
2
,
…
,
i
}
.
Prove
∣
A
n
∣
=
∑
k
=
0
∞
k
n
2
k
+
1
.
|\mathcal{A}_n| = \sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}}.
∣
A
n
∣
=
k
=
0
∑
∞
2
k
+
1
k
n
.
L. Lovasz
function
combinatorics proposed
combinatorics