Let the groups be group A,B,C,D The numbers in the 4 groups can be any of 5 ways: 2,2,2,6, that's (12C2)(10C2)(8C2)(6C6) = 66*45*28*1 = 83160 There are 4!/3! = 4 distinguishable permutations of that grouping. [BTW, FYI those 4 are 2,2,2,6; 2,2,6,2; 2,6,2,2; and 6,2,2,2.) So there are 83160*4 = 332640 for that grouping. 2,2,3,5, that's (12C2)(10C2)(8C3)(5C5) = 66*45*56*1 = 166320 There are 4!/2! = 12 distinguishable permutations of that grouping. So there are 166320*12 = 1995840 for that grouping. 2,2,4,4, that's (12C2)(10C2)(8C4)(4C4) = 66*45*70*1 = 173250 There are 4!/(2!2!) = 6 distinguishable permutations of that grouping. So there are 173250*6 = 1039500 for that grouping. 2,3,3,4, that's (12C2)(10C3)(7C3)(4C4) = 66*120*35*1 = 277200 There are 4!/(2!) = 12 distinguishable permutations of that grouping. So there are 277200*12 = 3326400 for that grouping. 3,3,3,3, that's (12C3)(9C3)(6C3)(3C3) = 220*84*20*1 = 369600 There are 4!/4! = only 1 permutation of that grouping. So there are 369600. Adding those together, 332640+1995840+1039500+3326400+369600 = 7063980 --------------------------------------------------------------- Edwin