m00li

Nicely done

continuing...

EDIT: the last term for F(n) should read n-[n/2]C[n/2]f(n-[n/2]) where [k] is the max integer <=k

I couldn't understand the question. every pair of odd squares and every pair of even square should qualify (e.g. 25-9=4*4, or 16-4=4*3)

there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black.
Method 1:
A = number of ways of selecting 5 out of the above
B = number of ways of selecting 5 such that no pawn is selected
C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected
then Answer = (A-C)/(A-B)
A = coeff of x5 in (1-x9)2(1-x3)6(1+x)4(1-x)-8 = 2540
B = coeff of x5 in (1-x3)6(1+x)4(1-x)-6 = 876
C = coeff of x5 in (1+x)(1-x9)(1-x3)6(1+x)4(1-x)-7=2230

Answer = 310/1664 = 0.186298

Method 2
A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected
B = no. of ways of selecting 5
C = no. of ways of selecting 5 without any pawns
Answer = A/(B-C)

B = (all 5 pieces are similar) + (4 similar and 1 different) + (3 sim, 2 sim) + (3 sim,2 different) + (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)
= (2C1) + (2C1*11C1) + (2C1*7C1) + (2C1*11C2) + (8C2*10) + (8C1*11C3) + (12C5) = 2540
C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)
= (6C2*8C1) + (6C1*9C3) + (10C5) = 876
A = (5 white pawn) + (4 WP, 1 different) + (3 WP, 2 similar pieces) + (3 WP, 2 different) + (2 WP, 2 sim,1 diff) + (2 WP, 3 diff)
= 1 + (11C1) + (7C1 + 11C2) + (7C1*10C1) + (11C3) = 310
Answer = 310/1664 = 0.186298