heres my try at this kind of problem:
we fix one HOD... the others can distribute in 29! ways... now we have to distribute the couples... i understand that the couples are fixed (ie to each woman theres a man assigned)... so they can arrange in 30! ways (as men have to sit on the right)... so anwer would be 29!*30!?
now, how would it be if the couples were not fixed (any man can sit behind any woman)?
we can make 15! couples, so answer would be 29!*15!*30!?
and finally, what if the positions of the couples were not fixed (ie man can sit either on the right or on the left)...
answer would be 29!*15!*30!*2^30?
s'one brighter than me please check my numbers... cheers, javi
If any man can sit next to any woman, assuming the men still sit to the same side next to each woman, the answer would be simply:
29! x 30! x 30!
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993