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Dom science forum beginner
Joined: 15 Feb 2006
Posts: 25

Posted: Sun May 07, 2006 9:01 am Post subject:
Re: How Many Combinations for THIS??



Proginoskes wrote:
Quote:  Dom wrote:
Proginoskes wrote:
Dom wrote:
Proginoskes wrote:
Dom wrote:
Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!
If the tiles are all indistinguishable, the answer is 4^16, since every
tile can be placed in one of 4 directions. If the tiles are all
distinct, the answer is 16! * 4^16 (16! = 1 * 2 * ... * 16), since
there are 16! ways to place 16 tiles.
So if I understand this right, if I have 16 different tiles (lets say
they are identified as A, B, C... P) there would be
89,862,698,310,039,500,000,000 different combinations?
That looks right.
However, would I be right in thinking that this could be divided by 4
because the whole grid will completed with an identical
sequence/pattern, but rotated through each 90 degrees?
Yes. And if you count horizontal and verticle reflections of a pattern
as being the same, you would divide by 2 (twice). But this wasn't
stated in the original problem.
Sorry, the rotation/reflection part didn't occur to me until after I had
originally posted!
So... dividing by 4 to remove rotation duplicates give me
22,465,674,577,509,900,000,000 combinations, then dividing by 2 to
remove reflection duplicates gives me 11,232,837,288,754,900,000,000
combinations so far. Excellent.
That only considers reflections in one direction to be the same
configuration. If neither reflection is considered to alter the tiling,
you divide by 2 again, to get 5,616,418,644,377,450,000,000.
Now to throw another spanner in the works! What if 8 out of the 16 tiles
couln't be placed in the centre 4 squares of the grid? How many
combinations would I be left with then?
I'm still assuming that this isn't homework.
Place the 8 tiles first: P(12, = 12*11*10*9*8*7*6*5, then place the
other 8, to get 8!. Now, rotating tiles is treated similarly:
P(12,*8!*4^16 total possibilities,
P(12,*8!*4^16/4 if rotations are considered to be the same
configuration,
P(12,*8!*4^16/16 if rotations and reflections are considered to be
the same configuration.
 Christopher Heckman

Hi Christopher,
Thanks for your explainations. It's not homework exactly  I found a
puzzle where the above rules apply and got it into my head that I wanted
to know how many combinations there could be. I've been trying to work
it out myself, but had no way of knowing whether my calculations were
correct (which they were up to a point, but then I just confused
myself). I don't know why it mattered to me really, but thanks for
putting my mind at ease!
Dom 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun May 07, 2006 6:05 am Post subject:
Re: How Many Combinations for THIS??



Dom wrote:
Quote:  Proginoskes wrote:
Dom wrote:
Proginoskes wrote:
Dom wrote:
Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!
If the tiles are all indistinguishable, the answer is 4^16, since every
tile can be placed in one of 4 directions. If the tiles are all
distinct, the answer is 16! * 4^16 (16! = 1 * 2 * ... * 16), since
there are 16! ways to place 16 tiles.
So if I understand this right, if I have 16 different tiles (lets say
they are identified as A, B, C... P) there would be
89,862,698,310,039,500,000,000 different combinations?
That looks right.
However, would I be right in thinking that this could be divided by 4
because the whole grid will completed with an identical
sequence/pattern, but rotated through each 90 degrees?
Yes. And if you count horizontal and verticle reflections of a pattern
as being the same, you would divide by 2 (twice). But this wasn't
stated in the original problem.
Sorry, the rotation/reflection part didn't occur to me until after I had
originally posted!
So... dividing by 4 to remove rotation duplicates give me
22,465,674,577,509,900,000,000 combinations, then dividing by 2 to
remove reflection duplicates gives me 11,232,837,288,754,900,000,000
combinations so far. Excellent.

That only considers reflections in one direction to be the same
configuration. If neither reflection is considered to alter the tiling,
you divide by 2 again, to get 5,616,418,644,377,450,000,000.
Quote:  Now to throw another spanner in the works! What if 8 out of the 16 tiles
couln't be placed in the centre 4 squares of the grid? How many
combinations would I be left with then?

I'm still assuming that this isn't homework.
Place the 8 tiles first: P(12, = 12*11*10*9*8*7*6*5, then place the
other 8, to get 8!. Now, rotating tiles is treated similarly:
P(12,*8!*4^16 total possibilities,
P(12,*8!*4^16/4 if rotations are considered to be the same
configuration,
P(12,*8!*4^16/16 if rotations and reflections are considered to be
the same configuration.
 Christopher Heckman 

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Dom science forum beginner
Joined: 15 Feb 2006
Posts: 25

Posted: Sat May 06, 2006 8:11 am Post subject:
Re: How Many Combinations for THIS??



Proginoskes wrote:
Quote:  Dom wrote:
Proginoskes wrote:
Dom wrote:
Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!
If the tiles are all indistinguishable, the answer is 4^16, since every
tile can be placed in one of 4 directions. If the tiles are all
distinct, the answer is 16! * 4^16 (16! = 1 * 2 * ... * 16), since
there are 16! ways to place 16 tiles.
So if I understand this right, if I have 16 different tiles (lets say
they are identified as A, B, C... P) there would be
89,862,698,310,039,500,000,000 different combinations?
That looks right.
However, would I be right in thinking that this could be divided by 4
because the whole grid will completed with an identical
sequence/pattern, but rotated through each 90 degrees?
Yes. And if you count horizontal and verticle reflections of a pattern
as being the same, you would divide by 2 (twice). But this wasn't
stated in the original problem.

Sorry, the rotation/reflection part didn't occur to me until after I had
originally posted!
So... dividing by 4 to remove rotation duplicates give me
22,465,674,577,509,900,000,000 combinations, then dividing by 2 to
remove reflection duplicates gives me 11,232,837,288,754,900,000,000
combinations so far. Excellent.
Now to throw another spanner in the works! What if 8 out of the 16 tiles
couln't be placed in the centre 4 squares of the grid? How many
combinations would I be left with then?
Thanks again for your advice in advance!
Dom 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sat May 06, 2006 5:47 am Post subject:
Re: How Many Combinations for THIS??



Dom wrote:
Quote:  Proginoskes wrote:
Dom wrote:
Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!
If the tiles are all indistinguishable, the answer is 4^16, since every
tile can be placed in one of 4 directions. If the tiles are all
distinct, the answer is 16! * 4^16 (16! = 1 * 2 * ... * 16), since
there are 16! ways to place 16 tiles.
So if I understand this right, if I have 16 different tiles (lets say
they are identified as A, B, C... P) there would be
89,862,698,310,039,500,000,000 different combinations?

That looks right.
Quote:  However, would I be right in thinking that this could be divided by 4
because the whole grid will completed with an identical
sequence/pattern, but rotated through each 90 degrees?

Yes. And if you count horizontal and verticle reflections of a pattern
as being the same, you would divide by 2 (twice). But this wasn't
stated in the original problem.
 Christopher Heckman 

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Dom science forum beginner
Joined: 15 Feb 2006
Posts: 25

Posted: Fri May 05, 2006 11:40 pm Post subject:
Re: How Many Combinations for THIS??



Proginoskes wrote:
Quote:  Dom wrote:
Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!
If the tiles are all indistinguishable, the answer is 4^16, since every
tile can be placed in one of 4 directions. If the tiles are all
distinct, the answer is 16! * 4^16 (16! = 1 * 2 * ... * 16), since
there are 16! ways to place 16 tiles.
 Christopher Heckman

So if I understand this right, if I have 16 different tiles (lets say
they are identified as A, B, C... P) there would be
89,862,698,310,039,500,000,000 different combinations?
However, would I be right in thinking that this could be divided by 4
because the whole grid will completed with an identical
sequence/pattern, but rotated through each 90 degrees? 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Fri May 05, 2006 10:43 pm Post subject:
Re: How Many Combinations for THIS??



Dom wrote:
Quote:  Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!

If the tiles are all indistinguishable, the answer is 4^16, since every
tile can be placed in one of 4 directions. If the tiles are all
distinct, the answer is 16! * 4^16 (16! = 1 * 2 * ... * 16), since
there are 16! ways to place 16 tiles.
 Christopher Heckman 

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Dom science forum beginner
Joined: 15 Feb 2006
Posts: 25

Posted: Fri May 05, 2006 10:10 pm Post subject:
How Many Combinations for THIS??



Okay...
If you have a 4x4 grid, with a "tile" in each square, how many placement
combinations are possible if any tile can be placed in any grid square
and can be orientated in any direction (in 90 degrees turns)?
If someone can explain how this would be calculated, this would be
GREATLY appreciated!!
Thanks for any advice in advance :)
Dom 

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