n factorial? wouldn't it be 2^(2n)?
Couldn't we reformulate the problem as "what proportion of 2n-bit integers have an equal number of 1 and 0 bits?"
& for n = 2:
BBBB, BBBG, BBGB, BBGG!, BGBB, BGBG!, BGGB!, BGGG, GBBB, GBBG!, GBGB!, GBGG, GGBB!, GGBG, GGGB, GGGG
P(G = B) = 4C2 / 2^4 = 6/16
not 6/24