HSC Stuff > HSC Mathematics Advanced
Round Table Probability
(1/1)
Jefferson :
Please help me with part (iii)
I thought about doing it like
Ways = [Harry, Meghan and Elizabeth all separated] + [Harry and Elizabeth together, Meghan separated]
= 2! * 3P3 + 2! * 3P2 * 2!
= 36
where the first 2! is the arrangement of the unnamed 3 people around a round table and the xPy is arranging H, M & E in between those unnamed people.
Are there any flaws in my way of thinking?
Also, are there any better approach to the problem? (e.g.using total ways, 5!, minus complement)
Thank you!
EDIT: I meant for this to be jn the 3U forums!! Apologies!
RuiAce:
Do you have the answer to this one? I might need an aim here because it's hard to track the exact correct method here.
Also, having trouble keeping track of where the numbers come from. Excluding H, M and E there are 5 more people right?
Jefferson :
--- Quote from: RuiAce on September 12, 2019, 10:17:56 pm ---Do you have the answer to this one? I might need an aim here because it's hard to track the exact correct method here.
Also, having trouble keeping track of where the numbers come from. Excluding H, M and E there are 5 more people right?
--- End quote ---
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 :(.
EDIT: Sorry, not sure what I was doing above. Was probably trying to do things without actually drawing a diagram. Please refer to my next comment for this approach (or attachment below).
RuiAce:
--- Quote from: Jefferson on September 15, 2019, 12:24:59 am ---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 :(.
--- End quote ---
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.
Jefferson :
--- Quote from: RuiAce on September 15, 2019, 01:11:10 pm ---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.
--- End quote ---
Hi RuiAce,
Sorry, I messed up my working twice in a row, but yes, my method also obtains 2400 ways as the answer.
Case 1
4! to arrange 5 people randomly, then 5P3 to arranged H, M, E in between them.
4! * 5P3 = 1440.
Case 2
4! to arrange 5 people randomly, then 5P2 * 2! to arrange H and E together as one group (can swap), and M separately.
4! * 5P2 * 2! = 960
TOTAL = 1440 + 960
TOTAL = 2400 ways
Doing it your way has given me a lot of insight into these type of questions. Since we both obtained the same answer (1 * 5 * 4 * 5! = 2400), I think this is what the solution is.
Again, Thank you!
-Jefferson
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