Hi RuiAce,
I don't have the solutions to these, unfortunately.
But you're absolutely right, there are 5 other guests. I counted it as 6 people by mistake, my bad!!
My working should look like:
Case 1: H, M, E all separated:
N1 = 4! * 6P3
Case 2: H & E together, M separated.
N2 = 4! * 3P2 * 2!
Hence, total ways is
NT = 4! * 6P3 + 4! * 3P2 * 2!
Solutions aren't provided sadly .
Although I'm not too sure with my own approach, I think how you did it is a bit dangerous. You started by arranging only the 5 other guests in a circle, giving you \(4!\) everywhere.
The thing with circular arrangements is that the \( (n-1)!\) technique applies only when you consider
every person, not just the unrestricted persons. Starting with this and then treating the question like a straight line arrangement (where you basically put objects between dividers) unwinds the reason why the \( (n-1)!\) trick works in the first place sadly!
Now of course in saying that I can't guarantee my approach works either. (Just because the restrictions are weirder.) But it's based on one of the two analogies of where the \((n-1)!\) comes from in the first place - deadlocking a person to create a starting/reference point. Fix the position of M. Then there are \(5\) positions for H, followed by \(4\) positions for E. And then the other guests.
This gives a total of \(5\times 4\times 5!\).
This seems to produce the same answer when I fix H instead. If I fix H, there are \(5\) positions for M. After drawing some diagrams, I found that no matter where I put \(M\), there will also be \(4\) places for E. And then the other guests, so again we have \(5\times 4\times 5!\).
I was going to also try deadlocking one of the 5 remaining guests instead, but I immediately saw that cases would then follow from that, which I wanted to avoid.