So given 4 elements, there are 15 partitions.
The formula for d is $d((x_1, x_2),(y_1, y_2)) = [(x_1−y_1)^2+ (x_2−y_2)^2]^{1/2}.$
I've calculated the distance for one of the partitions is $\{\{(0,1),(2,1)\},\{(3,1),(5,5)\}\}$.
The distance is $2 + (2)*(5^{1/2})$.
It was calculated by taking doing {(DISTANCE between (0,1),(2,1)) + (DISTANCE between (3,1),(5,5))}
and for another partition that has 4 singleton elements: $\{ \{(0,1)\}, \{(3,1)\}, \{(5,5)\}, \{(2,1)\} \}$
Intra-block distance for this partition would go like this: $$ d((0,1),(3,1)) +d((0,1,),(5,5)) +d((0,1),(2,1)) + d((3,1),(5,5))\\ + d((3,1),(2,1)) + d((5,5),(2,1))$$
My question is, for a partition that has 3 elements that has 2 singleton pairs and 1 element that has 2 coordinates like so: $\{\{(0,1),(2,1)\},\{(3,1)\},\{(5,5)\}\}$, how do I calculate the intrablock distance for this partition.