Methods for enumerating dimensional intersections

The 51st crease of the 6th to 9th dimensional vector.
Assuming there are 10 dimensions and we are assuming direct vector connection
between the 6th and 9th dimensions, as an edge of a 10-dimensional
hypercubic manifold, there are 10 minus 2 (the number of components in the
selected vector) factorial (for all permutations of intersection with
other 8 remaining dimensions by iterating through binary selection yeilds
the 51st datum of 8! items) 51 of 40320 with a lower bound of 0. 0 means
no intersections. The 51st binary permutation of the set {1, 2, 3, 4,
5, 7, 8, 10}. By choosing binary modality we get the enumeration 110011
or 32+16+3+1=51. The right-most binary numeral corresponds to the left-most
element of the set in one-to-one correspondance an intersection results in
the Turing subset {1, 2, 5, 7}.
The second method is iterative order, that is, to take all possible
single member sets (8 of them). Then 51-8=43. Then take all possible two
member sets which is 8*7 or 56. Since 43<56 therefore the 43rd member of
the set of two dimensional intersections, a set of sets: {{1, 2}, {1, 3},
..., {8,10}}. So the iterative answer is another Turing method:
1 {1,2}
2 {1,3}
3 {1,4}
4 {1,5}
5 {1,7}
6 {1,8}
7 {1,10}
This gets us to number eight so we can subtract n-1 from 43 until we get
a remainder and a quotient: 43-8=35, 35-7=28, 28-6=22, 22-5=16, 16-4=12,
12-3=9, 9-2=7, 7-1=6. So we result with the two-dimensional where the
members are the number of operations, the 8th, and the remainder, the 6
sixth. The result equals {A(6), A(8)} or {7, 10}. The 51st crease is an
intersection of the 6th, 7th, 8th and 10th dimensions.