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| Author |
Message |
Steel Dragon
science forum beginner
Joined: 11 Apr 2006
Posts: 2
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 Posted: Tue Apr 11, 2006 9:10 am Post subject:
"Simple" combinatory enigma - colored sticks in holes
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Hi,
A friend asked me to help her solve the following "simple"
problem, which is important to her. My skills in maths (I'm
a college dropout) are a bit rusty. I'd very much appreciate
if you people around could have a look at this and help me.
(Besides, forgive me in advance, but English is not my
mother tongue.) I have thought about a solution, but I'm
still working on it.
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
--
Steel Dragon |
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Jasen Betts
science forum Guru Wannabe
Joined: 31 Jul 2005
Posts: 176
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| Posted: Tue Apr 11, 2006 8:20 pm Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
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On 2006-04-11, Steel Dragon <nobody@somewhere_else.com> wrote:
| Quote: | Hi,
A friend asked me to help her solve the following "simple"
problem, which is important to her. My skills in maths (I'm
a college dropout) are a bit rusty. I'd very much appreciate
if you people around could have a look at this and help me.
(Besides, forgive me in advance, but English is not my
mother tongue.) I have thought about a solution, but I'm
still working on it.
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
|
since both stands are single-piece and gcf(8,3) = gcf(7,4) = 1
each hole can be identified uniquly by oservation)
therfore the stands themselves are not symmetrical althogh
the circles are, mirror symmetry is a possibility but not an issue here.
Draw maps and try rotating them if you don't believe me.
given that the answer to both the same as the number of permutations.
12! / 3!^4
Bye.
Jasen |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Wed Apr 12, 2006 1:51 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
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Steel Dragon wrote:
| Quote: | Hi,
A friend asked me to help her solve the following "simple"
problem, which is important to her. My skills in maths (I'm
a college dropout) are a bit rusty. I'd very much appreciate
if you people around could have a look at this and help me.
(Besides, forgive me in advance, but English is not my
mother tongue.) I have thought about a solution, but I'm
still working on it.
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
|
Don't feel bad ... this sort of problem isn't usually taught to people
who aren't mathematicians. (And you get bonus points for cross-posting
to alt.sci.math.combinatorics.)
If the center hole, one of the holes in the inner set of holes, and one
of the holes in the outer set of holes are all lined up, and the
reflection of an arrangement is considered to be the "same"
arrangement, then the answer to both problems is 12!/(2*3!*3!*3!*3!) =
184800. Otherwise, it is 12!/(3! 3! 3! 3!) = 369600.
The reason is: If there are 12 positions, and these positions are all
distinguishable, there are C(12,3) ways to place the red sticks, C(9,3)
ways to place the blue sticks, C(6,3) ways to place the green sticks,
and C(3,3) ways to place the yellow sticks. If you multiply these all
together, you get
C(12,3) * C(9,3) * C(6,3) * C(3,3) = 12! / (3! 3! 3! 3!).
If an arrangement and its reflection are to be counted as the same,
then you need to divide by 2.
--- Christopher Heckman |
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Steel Dragon
science forum beginner
Joined: 11 Apr 2006
Posts: 2
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| Posted: Wed Apr 12, 2006 8:19 pm Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
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Many thanks for both your answers, from me as well as
from my friend !
And it's reassuring that you come up to the same result.
I had had the idea to calculate what was (better written):
C(12,3) * C(9,3) * C(6,3) * C(3,3)
But first, I had done a basic mistake, and second, I was
not sure about not getting lost in the symetries of the
arrangement (or lack of symetry in this case, but my
friend had initially shown me slightly different setups).
--
Steel Dragon |
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S.
science forum beginner
Joined: 20 Apr 2006
Posts: 5
|
| Posted: Fri Apr 21, 2006 12:01 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
"Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1144806672.162374.193090@t31g2000cwb.googlegroups.com...
| Quote: |
Steel Dragon wrote:
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
If the center hole, one of the holes in the inner set of holes, and one
of the holes in the outer set of holes are all lined up, and the
reflection of an arrangement is considered to be the "same"
arrangement, then the answer to both problems is 12!/(2*3!*3!*3!*3!) =
184800. Otherwise, it is 12!/(3! 3! 3! 3!) = 369600.
The reason is: If there are 12 positions, and these positions are all
distinguishable, there are C(12,3) ways to place the red sticks, C(9,3)
ways to place the blue sticks, C(6,3) ways to place the green sticks,
and C(3,3) ways to place the yellow sticks. If you multiply these all
together, you get
C(12,3) * C(9,3) * C(6,3) * C(3,3) = 12! / (3! 3! 3! 3!).
If an arrangement and its reflection are to be counted as the same,
then you need to divide by 2.
--- Christopher Heckman
|
The stands are circular . Therefore they exhibit not just mirror symmetry
but
circular symmetry. E.g for a circular arrangement of n pieces/ sticks etc.
, n positions
( and not merely 2 ) can be made to coincide through rotation., and thus be
deemed equal,
As the 2 circular stands are differently constructed as regards the
distribution of positions
in the inner and outer circles, the number of possible arrangements
differs.
Stand (A) : 12! / (3! 3! 3! 3!) / 8 /3
Stand (B): 12! / (3! 3! 3! 3!) / 7 /4
BTW, if the statement of the problem is slightly alterted, such that the
number of all
such combinations is sought, when both stands are used together ( and not
seperately)
the problem becomes almost impossible to solve. Consider e.g. 2 red sticks
on Stand (A)
2 red sticks on Stand (B), similarly 3 red sticks on Stand (B) , 1 red
stick on Stand (A)
- each such seperation in turn yields "innumerable" seperations and
distributions of the
other sticks etc. etc.
This is not a simple problem by any stretch of imagination. Does any brave
soul wish
to try his hand at it ??? |
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Patrick Hamlyn
science forum beginner
Joined: 03 May 2005
Posts: 45
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| Posted: Fri Apr 21, 2006 12:16 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
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|
"S." <test@test.com> wrote:
| Quote: |
"Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1144806672.162374.193090@t31g2000cwb.googlegroups.com...
Steel Dragon wrote:
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
If the center hole, one of the holes in the inner set of holes, and one
of the holes in the outer set of holes are all lined up, and the
reflection of an arrangement is considered to be the "same"
arrangement, then the answer to both problems is 12!/(2*3!*3!*3!*3!) =
184800. Otherwise, it is 12!/(3! 3! 3! 3!) = 369600.
The reason is: If there are 12 positions, and these positions are all
distinguishable, there are C(12,3) ways to place the red sticks, C(9,3)
ways to place the blue sticks, C(6,3) ways to place the green sticks,
and C(3,3) ways to place the yellow sticks. If you multiply these all
together, you get
C(12,3) * C(9,3) * C(6,3) * C(3,3) = 12! / (3! 3! 3! 3!).
If an arrangement and its reflection are to be counted as the same,
then you need to divide by 2.
--- Christopher Heckman
The stands are circular . Therefore they exhibit not just mirror symmetry
but
circular symmetry. E.g for a circular arrangement of n pieces/ sticks etc.
, n positions
( and not merely 2 ) can be made to coincide through rotation., and thus be
deemed equal,
As the 2 circular stands are differently constructed as regards the
distribution of positions
in the inner and outer circles, the number of possible arrangements
differs.
Stand (A) : 12! / (3! 3! 3! 3!) / 8 /3
Stand (B): 12! / (3! 3! 3! 3!) / 7 /4
BTW, if the statement of the problem is slightly alterted, such that the
number of all
such combinations is sought, when both stands are used together ( and not
seperately)
the problem becomes almost impossible to solve. Consider e.g. 2 red sticks
on Stand (A)
2 red sticks on Stand (B), similarly 3 red sticks on Stand (B) , 1 red
stick on Stand (A)
- each such seperation in turn yields "innumerable" seperations and
distributions of the
other sticks etc. etc.
This is not a simple problem by any stretch of imagination. Does any brave
soul wish
to try his hand at it ???
|
I think you missed this bit:
| Quote: | Both stands are single-piece and circular shaped.
|
That means the circles can't be rotated. The stand can be rotated as a whole,
but not circles relative to each other.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms-subscribe@egroups.com) |
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Jasen Betts
science forum Guru Wannabe
Joined: 31 Jul 2005
Posts: 176
|
| Posted: Fri Apr 21, 2006 6:05 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
On 2006-04-21, S. <test@test.com> wrote:
| Quote: | The stands are circular . Therefore they exhibit not just mirror symmetry
but
circular symmetry. E.g for a circular arrangement of n pieces/ sticks etc.
, n positions
|
draw a map of the stands, they may have only mirror symmetry.
Bye.
Jasen |
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Jasen Betts
science forum Guru Wannabe
Joined: 31 Jul 2005
Posts: 176
|
| Posted: Fri Apr 21, 2006 9:54 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
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|
On 2006-04-21, S. <test@test.com> wrote:
| Quote: | The stands are circular . Therefore they exhibit not just mirror symmetry
but
circular symmetry. E.g for a circular arrangement of n pieces/ sticks etc.
, n positions
( and not merely 2 ) can be made to coincide through rotation., and thus be
deemed equal,
As the 2 circular stands are differently constructed as regards the
distribution of positions
in the inner and outer circles, the number of possible arrangements
differs.
Stand (A) : 12! / (3! 3! 3! 3!) / 8 /3
Stand (B): 12! / (3! 3! 3! 3!) / 7 /4
|
the thing is if the holes arent evenly spaced in each ring there's no
symmetry
if the hole are evenly spaced there'll be a pair of holes one from the inner
ring inner and one outer that is closer together than all other similar pairs.
that gives you a starting point to uniquely identify all the holes in both
rings.
(if two pairs of holes tie for closest there'll be a hole with a more
distant closet other-ring neighbours than all other holes and that can be a
strarting point.)
Bye.
Jasen |
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S.
science forum beginner
Joined: 20 Apr 2006
Posts: 5
|
| Posted: Sat Apr 22, 2006 2:01 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
"Jasen Betts" <jasen@free.net.nz> wrote in message
news:613f.4448abe1.6c5df@clunker.homenet...
| Quote: | On 2006-04-21, S. <test@test.com> wrote:
The stands are circular . Therefore they exhibit not just mirror symmetry
but
circular symmetry. E.g for a circular arrangement of n pieces/ sticks
etc.
, n positions
( and not merely 2 ) can be made to coincide through rotation., and thus
be
deemed equal,
As the 2 circular stands are differently constructed as regards the
distribution of positions
in the inner and outer circles, the number of possible arrangements
differs.
Stand (A) : 12! / (3! 3! 3! 3!) / 8 /3
Stand (B): 12! / (3! 3! 3! 3!) / 7 /4
the thing is if the holes arent evenly spaced in each ring there's no
symmetry
if the hole are evenly spaced there'll be a pair of holes one from the
inner
ring inner and one outer that is closer together than all other similar
pairs.
that gives you a starting point to uniquely identify all the holes in both
rings.
(if two pairs of holes tie for closest there'll be a hole with a more
distant closet other-ring neighbours than all other holes and that can be
a
strarting point.)
Bye.
Jasen
|
The devil's in the detail - the problem is ambiguous as regards the
construction of the stands.
However , given that the problem details the presence of 2 similar but
different stands with
a differing number of inner and outer rings, and thus one could expect,
differing results, I assumed
that the inner and outer rings are capable of being independently rotated.
Also, it is not impossible to have any number of evenly spaced holes along a
circular perimeter.
Under these assumptions, the solution is correct - right ?
Could someone venture (perhaps a ball-park figure) a guess as to
the number of different combinations,
(as asked in my previous message too) if
both stands are used together ( given that, as perhaps presumptiously
assumed previously,
the rings can be independently rotated and exhibit (to the extent of the
number of holes in them)
perfect rotational symmetry ?
Thanks. |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Sat Apr 22, 2006 5:32 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
S. wrote:
| Quote: | "Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1144806672.162374.193090@t31g2000cwb.googlegroups.com...
Steel Dragon wrote:
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
If the center hole, one of the holes in the inner set of holes, and one
of the holes in the outer set of holes are all lined up, and the
reflection of an arrangement is considered to be the "same"
arrangement, then the answer to both problems is 12!/(2*3!*3!*3!*3!) =
184800. Otherwise, it is 12!/(3! 3! 3! 3!) = 369600.
The reason is: If there are 12 positions, and these positions are all
distinguishable, there are C(12,3) ways to place the red sticks, C(9,3)
ways to place the blue sticks, C(6,3) ways to place the green sticks,
and C(3,3) ways to place the yellow sticks. If you multiply these all
together, you get
C(12,3) * C(9,3) * C(6,3) * C(3,3) = 12! / (3! 3! 3! 3!).
If an arrangement and its reflection are to be counted as the same,
then you need to divide by 2.
The stands are circular . Therefore they exhibit not just mirror symmetry but
circular symmetry. [...]
|
Normally, I would agree, but there are a different number of holes in
each circle. If you draw a picture and rotate it, you'll see that
rotating A by 45 degrees will make the holes in the outer circle line
up, but the ones in the inner one won't.
I considered this before I reponsed. 8-)
--- Christopher Heckman |
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|
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Sat Apr 22, 2006 5:33 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
S. wrote:
| Quote: | "Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1144806672.162374.193090@t31g2000cwb.googlegroups.com...
Steel Dragon wrote:
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
If the center hole, one of the holes in the inner set of holes, and one
of the holes in the outer set of holes are all lined up, and the
reflection of an arrangement is considered to be the "same"
arrangement, then the answer to both problems is 12!/(2*3!*3!*3!*3!) =
184800. Otherwise, it is 12!/(3! 3! 3! 3!) = 369600.
The reason is: If there are 12 positions, and these positions are all
distinguishable, there are C(12,3) ways to place the red sticks, C(9,3)
ways to place the blue sticks, C(6,3) ways to place the green sticks,
and C(3,3) ways to place the yellow sticks. If you multiply these all
together, you get
C(12,3) * C(9,3) * C(6,3) * C(3,3) = 12! / (3! 3! 3! 3!).
If an arrangement and its reflection are to be counted as the same,
then you need to divide by 2.
The stands are circular . Therefore they exhibit not just mirror symmetry but
circular symmetry. [...]
|
Normally, I would agree, but there are a different number of holes in
each circle. If you draw a picture and rotate it, you'll see that
rotating A by 45 degrees will make the holes in the outer circle line
up, but the ones in the inner one won't.
I considered this before I reponsed. 8-)
--- Christopher Heckman |
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| Back to top |
|
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Sat Apr 22, 2006 5:34 am Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
Proginoskes wrote:
| Quote: | S. wrote:
"Proginoskes" <CCHeckman@gmail.com> wrote in message
news:1144806672.162374.193090@t31g2000cwb.googlegroups.com...
Steel Dragon wrote:
There are 12 sticks, each being from one out of 4 possible
colors (red, blue, green and yellow - that is, 3 for each color
R, B, G, or Y).
The idea is to set them on twelve holes in a stand (for each
of a set of two different stands) and then to determine how
many different combinations we have.
Both stands are single-piece and circular shaped.
First stand (A) has one central hole plus two sets of holes, 8
in an outer circle, 3 in an inner circle.
Second stand (B) has one central hole plus two sets of holes,
7 in an outer circle, 4 in an inner circle.
Both stands are symmetrical - all holes will be evenly spaced
on each circle.
If the center hole, one of the holes in the inner set of holes, and one
of the holes in the outer set of holes are all lined up, and the
reflection of an arrangement is considered to be the "same"
arrangement, then the answer to both problems is 12!/(2*3!*3!*3!*3!) =
184800. Otherwise, it is 12!/(3! 3! 3! 3!) = 369600.
The reason is: If there are 12 positions, and these positions are all
distinguishable, there are C(12,3) ways to place the red sticks, C(9,3)
ways to place the blue sticks, C(6,3) ways to place the green sticks,
and C(3,3) ways to place the yellow sticks. If you multiply these all
together, you get
C(12,3) * C(9,3) * C(6,3) * C(3,3) = 12! / (3! 3! 3! 3!).
If an arrangement and its reflection are to be counted as the same,
then you need to divide by 2.
The stands are circular . Therefore they exhibit not just mirror symmetry but
circular symmetry. [...]
Normally, I would agree, but there are a different number of holes in
each circle. If you draw a picture and rotate it, you'll see that
rotating A by 45 degrees will make the holes in the outer circle line
up, but the ones in the inner one won't.
I considered this before I reponsed. 
|
Sorry for the double post --- Google Groups had a server error, and I
reposted.
--- Christopher Heckman |
|
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Jasen Betts
science forum Guru Wannabe
Joined: 31 Jul 2005
Posts: 176
|
| Posted: Sat Apr 22, 2006 1:05 pm Post subject:
Re: "Simple" combinatory enigma - colored sticks in holes
|
|
|
On 2006-04-22, S. <test@test.com> wrote:
| Quote: |
The devil's in the detail - the problem is ambiguous as regards the
construction of the stands.
However , given that the problem details the presence of 2 similar but
different stands with
a differing number of inner and outer rings, and thus one could expect,
differing results,
I assumed
that the inner and outer rings are capable of being independently rotated.
Also, it is not impossible to have any number of evenly spaced holes along a
circular perimeter.
|
the stands were specified as one-piece, a one-piece device doesn't have a
fixed and a rotating part.
| Quote: | Under these assumptions, the solution is correct - right ?
|
yes.
| Quote: | Could someone venture (perhaps a ball-park figure) a guess as to
the number of different combinations,
(as asked in my previous message too) if
both stands are used together ( given that, as perhaps presumptiously
assumed previously,
the rings can be independently rotated and exhibit (to the extent of the
number of holes in them)
perfect rotational symmetry ?
Thanks.
|
both stands (having rotatable rings) but how many sticks?
a single pool of sticks or one for each stand?
--
Bye.
Jasen |
|
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