Mixing
In how many ways can 24 items be split into 12 groups of 2 but with multiple rules applied to the grouping...
$begingroup$
I need to know a way to solve this question when there are rules applied to the grouping. For example 2 items of the same color can't be in a group or 2 items of the same size can't be in a group. Any match of a rule should rule the group out of the list of possibilities.
As requested per comment I will provide an actual use case. Suppose I have 24 wooden blocks. They can have different colors and can be of different sizes, let's say small, medium or large. They also have different shapes for example a cube, cylinder or pyramid.
So, suppose we have:
8 blue blocks: 3 small cubes, 3 medium cylinders, 2 large pyramids
8 red blocks: 3 small pyramids, 3 medium cylinders, 4 large pyramids
3 green blocks: 1 small cube, 1 large cube
3 purple blocks: 1 large pyramid, 1 large cube
2 yellow blocks: 2 large cubes
My question which I need to answer is how many groups of 2 (pairs) can there be made when none of the properties are the same. So you can't have 2 blue blocks as a group. You can't have 2 small blocks in a group. You can't have 2 cubes in a group. If a rule matches the pair is out.
Right now, I'm counting the distinct numbers of a rule (like 8 blue, 8 red, 3 green etc) and divide the total number of possibilities with 8!, 8! and 3! etc. However, somehow I think this is not totally correct ;)
I hope this makes sense and hope that someone could give me a big push in the right direction.
combinatorics
asked Jan 14 at 22:47
MartijnMartijn
62
$endgroup$
add a comment |
$begingroup$
I need to know a way to solve this question when there are rules applied to the grouping. For example 2 items of the same color can't be in a group or 2 items of the same size can't be in a group. Any match of a rule should rule the group out of the list of possibilities.
As requested per comment I will provide an actual use case. Suppose I have 24 wooden blocks. They can have different colors and can be of different sizes, let's say small, medium or large. They also have different shapes for example a cube, cylinder or pyramid.
So, suppose we have:
8 blue blocks: 3 small cubes, 3 medium cylinders, 2 large pyramids
8 red blocks: 3 small pyramids, 3 medium cylinders, 4 large pyramids
3 green blocks: 1 small cube, 1 large cube
3 purple blocks: 1 large pyramid, 1 large cube
2 yellow blocks: 2 large cubes
My question which I need to answer is how many groups of 2 (pairs) can there be made when none of the properties are the same. So you can't have 2 blue blocks as a group. You can't have 2 small blocks in a group. You can't have 2 cubes in a group. If a rule matches the pair is out.
Right now, I'm counting the distinct numbers of a rule (like 8 blue, 8 red, 3 green etc) and divide the total number of possibilities with 8!, 8! and 3! etc. However, somehow I think this is not totally correct ;)
I hope this makes sense and hope that someone could give me a big push in the right direction.
combinatorics
asked Jan 14 at 22:47
MartijnMartijn
62
$endgroup$
3
$begingroup$
Stating the actual problem would make your question clearer.
$endgroup$
– N. F. Taussig
Jan 14 at 22:52
$begingroup$
I have updated the original question with an acutal problem. Hope this will make things clearer.
$endgroup$
– Martijn
Jan 15 at 13:21
add a comment |
$begingroup$
I need to know a way to solve this question when there are rules applied to the grouping. For example 2 items of the same color can't be in a group or 2 items of the same size can't be in a group. Any match of a rule should rule the group out of the list of possibilities.
As requested per comment I will provide an actual use case. Suppose I have 24 wooden blocks. They can have different colors and can be of different sizes, let's say small, medium or large. They also have different shapes for example a cube, cylinder or pyramid.
So, suppose we have:
8 blue blocks: 3 small cubes, 3 medium cylinders, 2 large pyramids
8 red blocks: 3 small pyramids, 3 medium cylinders, 4 large pyramids
3 green blocks: 1 small cube, 1 large cube
3 purple blocks: 1 large pyramid, 1 large cube
2 yellow blocks: 2 large cubes
My question which I need to answer is how many groups of 2 (pairs) can there be made when none of the properties are the same. So you can't have 2 blue blocks as a group. You can't have 2 small blocks in a group. You can't have 2 cubes in a group. If a rule matches the pair is out.
Right now, I'm counting the distinct numbers of a rule (like 8 blue, 8 red, 3 green etc) and divide the total number of possibilities with 8!, 8! and 3! etc. However, somehow I think this is not totally correct ;)
I hope this makes sense and hope that someone could give me a big push in the right direction.
combinatorics
asked Jan 14 at 22:47
MartijnMartijn
62
$endgroup$
I need to know a way to solve this question when there are rules applied to the grouping. For example 2 items of the same color can't be in a group or 2 items of the same size can't be in a group. Any match of a rule should rule the group out of the list of possibilities.
As requested per comment I will provide an actual use case. Suppose I have 24 wooden blocks. They can have different colors and can be of different sizes, let's say small, medium or large. They also have different shapes for example a cube, cylinder or pyramid.
So, suppose we have:
8 blue blocks: 3 small cubes, 3 medium cylinders, 2 large pyramids
8 red blocks: 3 small pyramids, 3 medium cylinders, 4 large pyramids
3 green blocks: 1 small cube, 1 large cube
3 purple blocks: 1 large pyramid, 1 large cube
2 yellow blocks: 2 large cubes
My question which I need to answer is how many groups of 2 (pairs) can there be made when none of the properties are the same. So you can't have 2 blue blocks as a group. You can't have 2 small blocks in a group. You can't have 2 cubes in a group. If a rule matches the pair is out.
Right now, I'm counting the distinct numbers of a rule (like 8 blue, 8 red, 3 green etc) and divide the total number of possibilities with 8!, 8! and 3! etc. However, somehow I think this is not totally correct ;)
I hope this makes sense and hope that someone could give me a big push in the right direction.
combinatorics
combinatorics
asked Jan 14 at 22:47
MartijnMartijn
62
asked Jan 14 at 22:47
MartijnMartijn
62
edited Jan 15 at 13:20
Martijn
asked Jan 14 at 22:47
MartijnMartijn
62
asked Jan 14 at 22:47
MartijnMartijn
62
asked Jan 14 at 22:47
MartijnMartijn
62
62
3
$begingroup$
Stating the actual problem would make your question clearer.
$endgroup$
– N. F. Taussig
Jan 14 at 22:52
$begingroup$
I have updated the original question with an acutal problem. Hope this will make things clearer.
$endgroup$
– Martijn
Jan 15 at 13:21
add a comment |
3
$begingroup$
Stating the actual problem would make your question clearer.
$endgroup$
– N. F. Taussig
Jan 14 at 22:52
$begingroup$
I have updated the original question with an acutal problem. Hope this will make things clearer.
$endgroup$
– Martijn
Jan 15 at 13:21
3
3
$begingroup$
Stating the actual problem would make your question clearer.
$endgroup$
– N. F. Taussig
Jan 14 at 22:52
$begingroup$
Stating the actual problem would make your question clearer.
$endgroup$
– N. F. Taussig
Jan 14 at 22:52
$begingroup$
I have updated the original question with an acutal problem. Hope this will make things clearer.
$endgroup$
– Martijn
Jan 15 at 13:21
$begingroup$
I have updated the original question with an acutal problem. Hope this will make things clearer.
$endgroup$
– Martijn
Jan 15 at 13:21
add a comment |
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3
$begingroup$
Stating the actual problem would make your question clearer.
$endgroup$
– N. F. Taussig
Jan 14 at 22:52
$begingroup$
I have updated the original question with an acutal problem. Hope this will make things clearer.
$endgroup$
– Martijn
Jan 15 at 13:21