proofs

Cayley’s formula for the number of trees 205
Fourth proof (Double Counting). The following marvelous idea due
to Jim Pitman gives Cayley’s formula and its generalization (2) without
induction or bijection — it is just clever counting in two ways.
F 2 8 4
A rooted forest on {1, . . .,n} is a forest together with a choice of a root in
each component tree. Let F n,k be the set of all rooted forests that consist
of k rooted trees. Thus F n,1 is the set of all rooted trees.
3
5
Note that |F n,1 | = nT n , since in every tree there are n choices for the root.
2
We now regard F n,k ∈ F n,k as a directed graph with all edges directed 7 6
away from the roots. Say that a forest F contains another forest F ′ if F
contains F ′ as directed graph. Clearly, if F properly contains F ′ , then F
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has fewer components than F ′ 1
. The figure shows two such forests with the
roots on top. 10
Here is the crucial idea. Call a sequence F 1 , . . .,F k of forests a refining F 2 contains F 3
sequence if F i ∈ F n,i and F i contains F i+1 , for all i.
8 4 7
Now let F k be a fixed forest in F n,k and denote
• by N(F k ) the number of rooted trees containing F k , and
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• by N ∗ (F
2
1
k ) the number of refining sequences ending in F k .
F 3
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We count N ∗ (F k ) in two ways, first by starting at a tree and secondly by
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starting at F k . Suppose F 1 ∈ F n,1 contains F k . Since we may delete
the k − 1 edges of F 1 \F k in any possible order to get a refining sequence
from F 1 to F k , we find
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N ∗ (F k ) = N(F k )(k − 1)!. (3)
Let us now start at the other end. To produce from F k an F k−1 we have to
add a directed edge, from any vertex a, to any of the k −1 roots of the trees
that do not contain a (see the figure on the right, where we pass from F 3
to F 2 by adding the edge 3 7). Thus we have n(k − 1) choices.
Similarly, for F k−1 we may produce a directed edge from any vertex b to
any of the k−2 roots of the trees not containing b. For this we have n(k−2)
choices. Continuing this way, we arrive at
N ∗ (F k ) = n k−1 (k − 1)!, (4)
and out comes, with (3), the unexpectedly simple relation
N(F k ) = n k−1 for any F k ∈ F n,k .
For k = n, F n consists just of n isolated vertices. Hence N(F n ) counts the
number of all rooted trees, and we obtain |F n,1 | = n n−1 , and thus Cayley’s
formula.
□
But we get even more out of this proof. Formula (4) yields for k = n:
# { refining sequences (F 1 , F 2 , . . . , F n ) } = n n−1 (n − 1)!. (5)
For F k ∈F n,k , let N ∗∗ (F k ) denote the number of those refining sequences
F 1 , . . . , F n whose k-th term is F k . Clearly this is N ∗ (F k ) times the number
2
F 3 −→ F 2
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5
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