The maximum and minimum number of nodes in a binary tree of height 6 are :Note: A tree with zero node has height 0.A tree with one node has height 1.

A binary tree of height 'h' has a maximum and minimum number of nodes which can be calculated as follows:

1. Maximum number of nodes: A binary tree has the maximum number of nodes if all levels are fully filled. In a binary tree, the number of nodes at 'l' level is 2^(l-1). We know this because in a binary tree every node has at most 2 children. So in a binary tree of height 'h', the maximum number of nodes 'N' is the sum of the nodes of all 'h' levels.

The formula to calculate this is: N = 2^0 + 2^1 + 2^2 + ... + 2^(h-1)

This is a geometric progression with 'a' (first term) as 1, 'r' (common ratio) as 2 and 'n' (number of terms) as 'h'. The sum of 'n' terms of a geometric progression can be calculated as: N = a * (r^n - 1) / (r - 1)

Substituting the values we get: N = 1 * (2^h - 1) / (2 - 1) = 2^h - 1

So, the maximum number of nodes in a binary tree of height 6 is: N = 2^6 - 1 = 64 - 1 = 63

2. Minimum number of nodes: A binary tree has the minimum number of nodes if every level is not fully filled i.e., all nodes are added to the leftmost side of the previous level's last node. So in a binary tree of height 'h', the minimum number of nodes 'N' is 'h'.

So, the minimum number of nodes in a binary tree of height 6 is: N = 6

So, the maximum and minimum number of nodes in a binary tree of height 6 are 63 and 6 respectively.