hexagonal combination [on hold]
The figure below contains seven circles arranged in a hexagonal shape. Each circle is to contain a different integer between $1$ and $9$, inclusive, such that the sum of the three integers contained within any triangle is divisible by $3$. In how many ways can this be done? Two ways that differ only by rotation or reflection are distinguishable.
combinatorics
edited Jan 6 at 0:46
mrtaurho
4,06221133
asked Jan 6 at 0:09
weareallinweareallin
51
put on hold as off-topic by Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
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The figure below contains seven circles arranged in a hexagonal shape. Each circle is to contain a different integer between $1$ and $9$, inclusive, such that the sum of the three integers contained within any triangle is divisible by $3$. In how many ways can this be done? Two ways that differ only by rotation or reflection are distinguishable.
combinatorics
edited Jan 6 at 0:46
mrtaurho
4,06221133
asked Jan 6 at 0:09
weareallinweareallin
51
put on hold as off-topic by Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
The figure below contains seven circles arranged in a hexagonal shape. Each circle is to contain a different integer between $1$ and $9$, inclusive, such that the sum of the three integers contained within any triangle is divisible by $3$. In how many ways can this be done? Two ways that differ only by rotation or reflection are distinguishable.
combinatorics
edited Jan 6 at 0:46
mrtaurho
4,06221133
asked Jan 6 at 0:09
weareallinweareallin
51
The figure below contains seven circles arranged in a hexagonal shape. Each circle is to contain a different integer between $1$ and $9$, inclusive, such that the sum of the three integers contained within any triangle is divisible by $3$. In how many ways can this be done? Two ways that differ only by rotation or reflection are distinguishable.
combinatorics
combinatorics
edited Jan 6 at 0:46
mrtaurho
4,06221133
asked Jan 6 at 0:09
weareallinweareallin
51
edited Jan 6 at 0:46
mrtaurho
4,06221133
asked Jan 6 at 0:09
weareallinweareallin
51
edited Jan 6 at 0:46
mrtaurho
4,06221133
edited Jan 6 at 0:46
mrtaurho
4,06221133
edited Jan 6 at 0:46
mrtaurho
4,06221133
4,06221133
asked Jan 6 at 0:09
weareallinweareallin
51
asked Jan 6 at 0:09
weareallinweareallin
51
asked Jan 6 at 0:09
weareallinweareallin
51
51
put on hold as off-topic by Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, zipirovich, Leucippus, user91500, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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First let's fill it in with just the numbers $0,1,2$, which are the remainders of each number on division by $3$. Each triangle must have one of each. Put one of them in the center. You can't use any more of that remainder and have to alternate the other two remainders around the outside hexagon. That gives $3$ choices for the center and $2$ choices for the first cell, say the rightmost. There are $6$ ways to fill the diagram with remainders. Given the remainders, you can choose the center in $3$ ways and permute the outside cells in $(3!)^2$ ways, giving a total of
$$3 cdot 2 cdot 3 cdot (3!)^2=648$$ ways.
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First let's fill it in with just the numbers $0,1,2$, which are the remainders of each number on division by $3$. Each triangle must have one of each. Put one of them in the center. You can't use any more of that remainder and have to alternate the other two remainders around the outside hexagon. That gives $3$ choices for the center and $2$ choices for the first cell, say the rightmost. There are $6$ ways to fill the diagram with remainders. Given the remainders, you can choose the center in $3$ ways and permute the outside cells in $(3!)^2$ ways, giving a total of
$$3 cdot 2 cdot 3 cdot (3!)^2=648$$ ways.
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
add a comment |
First let's fill it in with just the numbers $0,1,2$, which are the remainders of each number on division by $3$. Each triangle must have one of each. Put one of them in the center. You can't use any more of that remainder and have to alternate the other two remainders around the outside hexagon. That gives $3$ choices for the center and $2$ choices for the first cell, say the rightmost. There are $6$ ways to fill the diagram with remainders. Given the remainders, you can choose the center in $3$ ways and permute the outside cells in $(3!)^2$ ways, giving a total of
$$3 cdot 2 cdot 3 cdot (3!)^2=648$$ ways.
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
add a comment |
First let's fill it in with just the numbers $0,1,2$, which are the remainders of each number on division by $3$. Each triangle must have one of each. Put one of them in the center. You can't use any more of that remainder and have to alternate the other two remainders around the outside hexagon. That gives $3$ choices for the center and $2$ choices for the first cell, say the rightmost. There are $6$ ways to fill the diagram with remainders. Given the remainders, you can choose the center in $3$ ways and permute the outside cells in $(3!)^2$ ways, giving a total of
$$3 cdot 2 cdot 3 cdot (3!)^2=648$$ ways.
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
First let's fill it in with just the numbers $0,1,2$, which are the remainders of each number on division by $3$. Each triangle must have one of each. Put one of them in the center. You can't use any more of that remainder and have to alternate the other two remainders around the outside hexagon. That gives $3$ choices for the center and $2$ choices for the first cell, say the rightmost. There are $6$ ways to fill the diagram with remainders. Given the remainders, you can choose the center in $3$ ways and permute the outside cells in $(3!)^2$ ways, giving a total of
$$3 cdot 2 cdot 3 cdot (3!)^2=648$$ ways.
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
answered Jan 6 at 0:19
Ross MillikanRoss Millikan
292k23197371
292k23197371
add a comment |
add a comment |
