Our old friend STAN.K bought n pairs of white "chiral" (podal!) socks clearly stamped L and R, which his housekeeper nevertheless paired willy-nilly. Stan wondered, for n pairs, how likely were 0, 2, 4, ... mismatches. I couldn't remember how to map a washer-dryer onto an urn, so I just tabulated 2 4 2 8 12 16 48 6 32 160 60 64 480 360 20 128 1344 1680 280 256 3584 6720 2240 70 512 9216 24192 13440 1260 and fit a formula with Mathematica. The nth row represents n pairs, row sum = binomial(2n,2). Thus the 2nd row, 4 2, represents 2 pairs, LRLR,LRRL,RLLR,RLRL = 4 perfectos, and LLRR,RRLL = 2 mismatches. (The left column is how many perfectos for n pairs.) The empirical formula I got was StringReverse@"]j2-n,j,j[laimonitluM)j2-n(^2" which then Julian promptly derived in a single line: StringReverse@".LR dna RL neewteb esoohc ot syaw )j2-n(^2 eht yb deilpitlum neht si hcihw ,)!)j2-n(*2^!j(/!n si LR ro RL rehtie eb ot j2-n dna ,RR eb ot j ,LL eb ot sriap j esoohc ot syaw fo rebmun eht" I guess that makes it fairly clear why the left (perfecto) column is 2^n. And, why the right diagonal is 2n choose n. --rwg