Great approach, shankar.ashwin!shankar.ashwin wrote:Well, I think circular permutation (n-1)! works only when there is only 1 element present. Here since there are 3 entities, treat the problem no different than a normal permutation problem.
In basic arrangements,
_ _ _ (Considering the first set of 3 to be seated)
Any of the 30 can seat in the first (M, W or C)
The second slot can be filled by 20 possibilites (Omitting M,W or C who have already seated in the first)
The third can be seated by any of the remaining 10 (again excluding 2 of M,W or C who have already been seated)
So, we have 30*20*10 for the first 3.
Here you have already accounted for order, so all of the remaining 9 M,W or C can fill up 9 slots in 9!*9!*9! ways..
Combining we have 6000* 9!*9!*9! ways. (which is (6)(10!)(10!)(10!) )
I think this is pretty straightforward. (Circular arrangements between 3 different elements does not make sense)
However, if we've agreed that location of the seats doesn't matter, then we need to take your answer of (6)(10!)(10!)(10!) and divide it by 30 to eliminate all of the duplication (see above for rationale).
Once we divide by 30, we get (10!)(10!)(10!)/5
Cheers,
Brent
