knight247 wrote:In how many different ways can you seat 10 women, 10 men, and 10 children around a circular table if you need to alternate the men, women and children?
I'm pretty sure this question is beyond the scope of the GMAT, but here's my attempt.
First, let's pretend that the location of the seats
does matter. This means that, if we have one arrangement, and then everyone shifts 1 seat to the left, this creates a new arrangement, even though the relative location of each person is the same.
What are the implications here? We will be counting each arrangement more than once. In fact, we will be counting each arrangement 30 times, rather than just 1 time.
Why?
Well, if we take one arrangement, and then have everyone shift one seat to the left to get a "different" arrangement, we can keep doing this until each person has occupied all 30 seat.
So, if we assume that each that the location of the seat
does matter, then the number of arrangements we calculate must later be divided by 30 to eliminate all of the duplication.
Okay, at this point, we can number the chairs #1, #2, #3 and so on.
We have 2 cases to consider:
case 1) the order is MWC MWC MWC etc
case 2) the order is MCW MCW MCW etc
Case 1
Let's place a man is seat #1. This means that the men will occupy seats, #1, 4, 7, 10,...
We can seat a man in seat #1 in 10 ways, then seat a man in seat #2 in 9 ways, etc.
So, we can seat the men in 10! ways.
Next we seat the women. They will occupy seats #2, 5, 8, 11, ...
We can seat a woman in seat #2 in 10 ways, then seat a woman in seat #5 in 9 ways, etc.
So, we can seat the woman in 10! ways.
Next we seat the children. They will occupy seats #3, 6, 9, 12,...
We can seat a child in seat #3 in 10 ways, then seat a child in seat #6 in 9 ways, etc.
So, we can seat the children in 10! ways.
So, we can seat all 30 people in (10!)(10!)(10!) ways [assuming that the location of each seat matters]
Important: Before we examine case 2, we should also recognize that I could have also seated the men in seats #2, 5, 8, 11,..., which means the women would have been seated in seats #3, 6, 9, 12,... and the children in seats #4, 7, 10, ...
So, if we seat the men in seats #2, 5, 8, 11,... (and then seat the women and children), we can accomplish this in (10!)(10!)(10!) ways [by applying the same logic as earlier]
Also, we can seat the men in seats #3, 6, 9, 12,..., which means the women would have been seated in seats #4, 7, 10, ... and the children in seats #5, 8, 11, ...
This can be accomplished in (10!)(10!)(10!) ways as well.
At this point, we have considered all of the possible MWC configurations.
So, for case 1, the total number of ways to seat all 30 people = (10!)(10!)(10!)+(10!)(10!)(10!)+(10!)(10!)(10!) = (3)(10!)(10!)(10!)
Case 2
Using the logic we used for case 1, we can seat all 30 people in (3)(10!)(10!)(10!) ways
Combining both case, we see that we can seat everyone in (6)(10!)(10!)(10!) ways [assuming that the location of each seat
matters]
At this point, we can account for duplication by dividing by 30 to get:
(6)(10!)(10!)(10!)/30, which equals
(10!)(10!)(10!)/5 ways
Proviso: I'm not feeling 100% about this solution - perhaps this morning's coffee hasn't kicked in yet
Cheers,
Brent
