Originally posted by SwissGambit
This is more a math problem than a chess problem. I'll start by doing the 'chess' legwork, and maybe some math genius will fill in the rest.
One of the SPGs leading to the diagram is:
1. b4 a5 2. e4 a4 3. Bb2 Ra5 4. Ne2 Rf5 5. Ng3 e5 6. Be2 b5 7. O-O Qg5 8.
Re1 Bb7 9. Nf1 Bd5 10. g3 Be6 11. Kg2 Bd6 12. Bd4 Nf6 13. Kh3 Kd8 14. Be3
13.5 moves appea ...[text shortened]... ced automatically, and thus do not generate any 'extra' combinations.
more to come...
So, to determine the number of white move orders...
Label the sequence b4-Bb2-Bd4-Be3 Q1, Q2, Q3, Q4 respectively. Start with Q4 as move 14 and Q3 as move 13. This leaves 11 different places to put Q2. If Q2 is move 12, that leads to 11 places to put Q1 and thus 11 move orders. If Q2 is move 11, we get 10 move orders and so on. 13.Q3 and 14.Q4 leads to 1+2+3+4+5+6+7+8+9+10+11 = 66 move orders.
With 12.Q3 and 14.Q4, there is one less place to stick Q2/Q1, so we only get 1+2+3+4+5+6+7+8+9+10 = 55 move orders.
Each time I shift Q3 one move left [sooner in the game], I end up cutting a number off the addition stack. Spreadsheeting this, I get 66+55+45+36+28+21+15+10+6+3+1 = 286 move orders.