Pawn Structure challenge

First draft:

[Pawn a may be in any of the positions [a-f]×[2-7], a total of 36 positions. Pawn b may be in [a-g]×[2-7] = 42 positions, minus the one position already occupied by pawn a. Pawns c - f all have 48 legal positions, and must avoid each other and the pawns a-b so for pawns a to f we have 36×41×46×45×44×43 positions. For pawns g and h it becomes a bit more complicated depending on where exactly the previous pawns live and we must introduce a bit more math.

We partition the colums in P = {a, b, center, g, h} and may attach to each pawn configuration a map F: P -> (integer functions on P), saying how many pawns went from here to there. The map must obviously satisfy: F(a) has weight one and is supported only by {a,b,center}, F(b) has weight one and is supported only by {a,b,center,g} etc. And of course, the 7 pawns a-g may all go to column b, which however has only 6 places for them, so this is an extra restriction. This allows us to enumerate the set of allowed functions F, which I am not going to do.

Now we can redo the computations above for a particular function F. As an example I assume that: F(a) = {a: 1}, F(b) = {center: 1}, F(center) = {b:1, center:2, g: 1}, F(g) = {g:1}, F(h) = {h:1} (with obvious notations). Writing A(n,m) = n!/(n-m)! for the number of injections of m->n, this gives us: for column a, A(6,1) positions; for the b-pawn gone to center, 36 positions; for the center pawn gone to b column, C(4,1)=4 choices of pawn times A(6,1) = 6 possible positions; for the two center pawns gone to center, C(4-1,2) choices of pawn (excluding the center->b pawn) times A(36-1,2) possible positions (excluding the spot taken by the b-pawn); for the center->g pawn, C(4-3,1) times A(6,1) positions; for the g->g pawn, C(1,1) times A(6-1,1); for the h->h pawn, C(1,1) times A(6,1). The product is ugly.](#sp)

We may still be able to find a nicer formula for the total, but the more interesting question would be: what about unlabeled pawn structures?

/r/mathriddles Thread