I have another permutation question that I'm having trouble with; this time with circular arrangements:
To a meeting involving four companies, each company sends three representatives -- the managing director, the chief accountant and the company secretary. In how many ways can the twelve people be arranged around a circular table if the three people from each company sit together, with the managing director between the accountant and the secretary in each case?
I know that in circular arrangements, the possible number of permutations is $(n-1)!$ - so in this case, we have $3! = 6$. The answer listed, however, is 96. Where am I going wrong?
We have four groups of three arranged around the circle. First choose the four seats of the managing directors. This can be done in $3$ ways. Next, distribute the four companies around the four quarters of the circle (around the four chosen seats). That can be done in $4!=24$ ways. Next, for each company, choose who sits to the left an who sits to the right of the managing director. This can be done in $2^4=16$ ways. Together, we get $$ 3\cdot 24\cdot 16=1152 $$ possibilities. Because of the circular arrangement, we have to divide by $12$ to get $96$.