[QUOTE=xilman][b]Now[/b] do you see why the riddle is ambiguous? My friends met a man with three brothers. They did not meet all of my brothers.[/QUOTE]
Indeed, they met two men with three brothers each. How many brothers were there in all?


Paul

A riddle rhyme.
 

[QUOTE=xilman]Indeed, they met two men with three brothers each. How many brothers were there in all?
Paul[/QUOTE]
:smile:
Paul,
Compliments of the season.
Your problem is as ambiguous as mine!
Are the two men, brothers themselves? If not then the answer is twelve.
If they are then the answer is six.
Thats if we don't consider the two men who met them. If we do, then are there any takers? :question:
Mally :coffee:

[QUOTE=mfgoode]:smile:
Paul,
Compliments of the season.
Your problem is as ambiguous as mine!
Are the two men, brothers themselves? If not then the answer is twelve.
If they are then the answer is six.
[/QUOTE]


Maybe we can consider the fact that the two are brothers and have another two away -- that would give the soltuion of two (more ambiguity)

P.S: If the two men are brothers we can even achieve a solution of four, if one brother is allowed to count for both men.

[QUOTE=fetofs]P.S: If the two men are brothers we can even achieve a solution of four, if one brother is allowed to count for both men.[/QUOTE]
Four is the correct answer in this case.

I have three brothers (Ian, Neil and Nigel). Nigel has three brothers (me, Ian and Neil).

Nigel visited me. Ian and Neil did not. Two brothers were present, each of whom have three brothers.


Paul

[QUOTE=mfgoode]:smile:
Paul,
Compliments of the season.
Your problem is as ambiguous as mine!
Are the two men, brothers themselves? If not then the answer is twelve.
If they are then the answer is six.
Thats if we don't consider the two men who met them. If we do, then are there any takers? :question:
Mally :coffee:[/QUOTE]
Is the wrong answer.

Re-read the description of the party I held.


Paul

[QUOTE=xilman]Four is the correct answer in this case.

I have three brothers (Ian, Neil and Nigel). Nigel has three brothers (me, Ian and Neil).

Nigel visited me. Ian and Neil did not. Two brothers were present, each of whom have three brothers.


Paul[/QUOTE]

I had misinterpreted the question when I gave my first answer; I said "two" because that was the number of brothers "present" on the party. :wink:

A Riddle rhyme
 

[QUOTE=Wacky]Or perhaps the question asks you to avoid counting the men at all.

As pointed out, the problem is quite ambiguous.[/QUOTE]
:surprised
Thats lateral thinking Wacky. Its an angle I didn't consider at all!
Well ,the very fact that I gave a hint of two answers, is sufficient to conclude that the problem is ambiguous and I don't deny it in the least. Its interesting to see the various interpretations of a school child's rhyme.
Mally :coffee:

[QUOTE=tom11784]I say there are 3 answers:
[spoiler]1 - just me (as seen in "Die Hard With a Vengence")[/spoiler]
[spoiler]2 - me and the man (who has the wives, but left them home)[/spoiler]
[spoiler]2753 - me and the man (who brought the 7 wives, 343 cats, and 2401 kits - sacks don't count)[/spoiler][/QUOTE]
:unsure: How do you arrive at the figure of 2753 ? Is it a typo error?
Mally :coffee:

A Riddle rhyme
 

[QUOTE=xilman]Indeed, they met two men with three brothers each. How many brothers were there in all?


Paul[/QUOTE]
:smile: Well lets consider the algebraic solution.
Lets forget the 'they'
We have two men (A and B) with three brothers each.
If the men themselves are not brothers but unrelated we have two sets A and B
Each set (A or B ) consists of 4 elements (brothers).
If we take a subset of two elements (brothers) then we have a subset of 2 brothers
So from A we derive 4C2 brothers = 4*3/1*2 = 6
The same goes for set B = 6
In all (A and B) we have 6+6 brothers = 12 brothers (sub sets) :rolleyes:
If the two men are themselves brothers then the set consists of 4 elements
Similarly there are 6 subsets. and so 6 sets of brothers. :rolleyes:
Elementary my dear Watson!
Mally :coffee: .

[QUOTE=mfgoode]
Elementary my dear Watson!
Mally :coffee: .[/QUOTE]

Your reasoning is incorrect, because if the men are brothers, a brother of one of them would be a brother of the other one as well. That's elementary. In your solution, you only consider one men to count for the other.

By the way, for the basic solution, I can count 8 (Set A+Set B). How did four brothers appear magically?

Edit: Oh, I see. You're not counting the [I]elements[/I], you're counting the [I]subsets[/I]. Indeed, there are 6 subsets: 1,2,3,1 and 2, 2 and 3, 1 and 3. But the problem isn't asking for the number of subsets you can form with 3 brothers, is it?

:smile:
Thank you fetofs. It shows you are thinking and that's good. I'm not infallible you know but I like an intelligent discussion and welcome your viewpoint.

Lets put it this way. Let one set of brothers be called X and the other Y.
Lets say X consists of four brothers A, B, C, D. The minimum we can have to call them a single group of brothers is a combination of two of them.

We thus have AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC. Obviously AB and BA are the same brothers so are AC and CA and so on and so we eliminate the duplicates.

We thus get remaining AB, AC, AD, BC, BD, CD. = 6 groups of brothers.
The same goes for the set Y = 6 groups of brothers.
Algebraically this is written as 4 C 2 where C is a combination = 4*3/1*2
We can generalise this formula for any number of brothers N as
NCr = N!/(N-2)!*2. which is the minimum combination for a group of brothers taken two at a time to make the group. :confused: :question:
Mally :coffee: