The number of different trees with 8 nodes is
2011
The number of different trees with 8 nodes is
Answer: D. None of these — Counting "different trees" on a fixed number of nodes is well defined only once the counting convention is fixed. Three independent choices set the total:…
- A.
256
- B.
255
- C.
248
- D.
None of these
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Correct answer: D
Counting "different trees" on a fixed number of nodes is well defined only once the counting convention is fixed. Three independent choices set the total: whether the nodes carry distinct labels, whether one node is singled out as a root, and whether the order of a node’s children matters. Changing any one of them changes the count, so a claimed tree total has to be checked against every standard convention, not just one.
Non-isomorphic (unlabelled) free trees — nodes are interchangeable and no root is fixed, so two trees are the same whenever one can be redrawn as the other; the totals are tabulated (OEIS A000055) and have no simple closed formula.
Labelled free trees — the n nodes carry distinct labels, and Cayley’s formula gives nn − 2.
Rooted unlabelled trees — one node is singled out as the root while a node’s children stay unordered; these totals are tabulated too (OEIS A000081).
Ordered binary trees — every node has a distinguishable left and right child slot, and the total is the Catalan number Cn = (1/(n + 1)) × C(2n, n).
Apply each convention at n = 8:
Unlabelled free trees on 8 nodes: 23. Harary’s Graph Theory states outright that there are 23 different trees with eight points (OEIS A000055).
Labelled free trees on 8 nodes: Cayley’s formula gives 86 = 262144.
Rooted unlabelled trees on 8 nodes: 115 (OEIS A000081).
Ordered binary trees with 8 nodes: the Catalan number C8 = (1/9) × C(16, 8) = 12870/9 = 1430.
Now set the listed values beside those four totals — 23, 115, 1430 and 262144:
Listed value | What that number is | A standard tree total? |
|---|---|---|
256 | 28, the count of 8-bit binary strings | No |
255 | 28 − 1, the largest 8-bit unsigned integer | No |
248 | 28 − 8, that is 2n − n taken at n = 8 | No |
Some circulated keys reach 248 by applying "2n − n". That expression is not a tree-enumeration formula and it already breaks at the first small case: at n = 4 it gives 12, whereas there are 2 unlabelled trees, 16 labelled trees (42 by Cayley’s formula) and 14 ordered binary trees on 4 nodes.
No standard convention yields 256, 255 or 248 for 8 nodes, so the correct choice is "None of these".