Can some body help me to find total number of nodes in this tree network see the picture tree network, here Level-1 one has 4 nodes, and the next level has 3 and so 2, 2 is the minimum number of children a parent can have and there are maximum 3 levels allowed. What is the mathematical way of writing this in a summation form. for example the number of nodes in this example is 41 but how such a number is calculated through formula. If the nodes in the levels are not known then what would be its general formula?
Max number of nodes in non-binary tree in summation form
1k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
This post explains how to calculate an estimation of the total number of nodes in the tree, not the exact number.
If n is the number of levels in a similar tree, the number of nodes in the final level is n! (n factorial). The number of nodes in all the levels working backwards is as follows:
- Level - Total Nodes
- n - n! (same as saying n!/1!)
- n-1 - n!/2! (total nodes in level n divided by 2)
- n-2 - n!/3! (total nodes in level n divided by 2, then by 3)
- n-3 - n!/4! (total nodes in level n divided by 2, then by 3, then by 4)
- ...
- root node - n!/n! (equals 1)
If you add up the formulas for total nodes at each level and substitute n!/1! for n! since they are equal, you would get the following formula:
n!/1! + n!/2! + n!/3! + n!/4! + ... + n!/n!
take out the common factor n!
n!(1/1! + 1/2! + 1/3! + 1/4! + ... + 1/n!)
the stuff in the parentheses is similar to the following formula for the constant e:
e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ...
except it's missing the initial number 1, therefore the formula to approximate the total number of nodes in the entire tree is
n!(e-1) where n is the number of levels (depth) in the tree.
If you apply this formula to the tree in the original question, there are 4 levels so 4!(e-1) is about 41.2387, and the actual number of nodes is 41.

Your problem can be simply described with the following equation:
$N=\sum_{n=1}^{N}{\frac{N!}{(N-n)!}}= \sum_{n=1}^{4}{\frac{4!}{(4-n)!}}$
With N-1 being the maximum level of 3. i.e. N=4.