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Currently I am reading a paper where the author states:

[...] It is well-known that an infinite $D$-regular rooted tree contains precisely $\frac{1}{(D-1)u + 1} \binom{Du}{u}$ rooted subtrees of size $u$ [...]

Unfortunately this is not "well-known" to me, not being an expert in combinatorics. I understand this in the following sense: Starting from the root how many different subtrees can one construct containing $u$ sites. Or am I already wrong here? Is it possible to obtain the result based on a recursion relation? I am happy to work with a hint instead of an direct answer/proof...

antarcticfox
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  • In the meantime I was able to show it for an infinite $D-ary$ tree, i.e. every node has $D$ children, by using the generating function method. But in a $D-regular$ tree all nodes, except the root, have $D-1$ children... – antarcticfox May 21 '14 at 07:00
  • I don't understand the definition of $D$-regular tree. It sounded trivial to me that if the tree is $D$-regular, then the root must have $D$ children. And, any other node must have $D - 1$ children to be of degree $D$, as it must have exactly one parent. Why do we need generating functions ? I also don't understand the rooted subtrees of size $u$. Maybe you can tell me what's wrong with this : Say you pick any non-root node $x$. Take the subtree rooted at $x$, and pick $u$ nodes in this subtree. You got a subtree of size $u$. So there's an infinity of them. – Manuel Lafond May 21 '14 at 17:08
  • Starting from the root how many ways are there to choose a sub-tree with u vertices (including the root). The statement that the tree is infinite is only there to assure that we can do so for any $u$ we like. If we fix the root there is obviously only a finite number of possibilities for finite $u$. – antarcticfox May 21 '14 at 17:21
  • Ah, so the root of the subtrees we count must be the same root as the tree... – Manuel Lafond May 21 '14 at 17:25

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