Mixing
Nine objects in non-empty boxes [closed]
$begingroup$
In how many ways 9 identical objects can be put in non-empty boxes of arbitrary size?
Is solution integer partition of 9? That is 30?
combinatorics integer-partitions
asked Jan 14 at 16:27
StratocarterStratocarter
146
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closed as off-topic by Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin Jan 17 at 12:23
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." – Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In how many ways 9 identical objects can be put in non-empty boxes of arbitrary size?
Is solution integer partition of 9? That is 30?
combinatorics integer-partitions
asked Jan 14 at 16:27
StratocarterStratocarter
146
$endgroup$
closed as off-topic by Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin Jan 17 at 12:23
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." – Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second?
$endgroup$
– saulspatz
Jan 14 at 16:29
$begingroup$
The boxes are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 16:33
$begingroup$
Then it's partitions, as you said. (I haven't checked that the answer is $30$.)
$endgroup$
– saulspatz
Jan 14 at 16:34
$begingroup$
Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second.
$endgroup$
– Stratocarter
Jan 14 at 16:34
add a comment |
$begingroup$
In how many ways 9 identical objects can be put in non-empty boxes of arbitrary size?
Is solution integer partition of 9? That is 30?
combinatorics integer-partitions
asked Jan 14 at 16:27
StratocarterStratocarter
146
$endgroup$
In how many ways 9 identical objects can be put in non-empty boxes of arbitrary size?
Is solution integer partition of 9? That is 30?
combinatorics integer-partitions
combinatorics integer-partitions
asked Jan 14 at 16:27
StratocarterStratocarter
146
asked Jan 14 at 16:27
StratocarterStratocarter
146
asked Jan 14 at 16:27
StratocarterStratocarter
146
asked Jan 14 at 16:27
StratocarterStratocarter
146
asked Jan 14 at 16:27
StratocarterStratocarter
146
146
closed as off-topic by Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin Jan 17 at 12:23
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." – Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin Jan 17 at 12:23
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." – Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second?
$endgroup$
– saulspatz
Jan 14 at 16:29
$begingroup$
The boxes are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 16:33
$begingroup$
Then it's partitions, as you said. (I haven't checked that the answer is $30$.)
$endgroup$
– saulspatz
Jan 14 at 16:34
$begingroup$
Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second.
$endgroup$
– Stratocarter
Jan 14 at 16:34
add a comment |
1
$begingroup$
Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second?
$endgroup$
– saulspatz
Jan 14 at 16:29
$begingroup$
The boxes are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 16:33
$begingroup$
Then it's partitions, as you said. (I haven't checked that the answer is $30$.)
$endgroup$
– saulspatz
Jan 14 at 16:34
$begingroup$
Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second.
$endgroup$
– Stratocarter
Jan 14 at 16:34
1
1
$begingroup$
Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second?
$endgroup$
– saulspatz
Jan 14 at 16:29
$begingroup$
Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second?
$endgroup$
– saulspatz
Jan 14 at 16:29
$begingroup$
The boxes are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 16:33
$begingroup$
The boxes are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 16:33
$begingroup$
Then it's partitions, as you said. (I haven't checked that the answer is $30$.)
$endgroup$
– saulspatz
Jan 14 at 16:34
$begingroup$
Then it's partitions, as you said. (I haven't checked that the answer is $30$.)
$endgroup$
– saulspatz
Jan 14 at 16:34
$begingroup$
Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second.
$endgroup$
– Stratocarter
Jan 14 at 16:34
$begingroup$
Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second.
$endgroup$
– Stratocarter
Jan 14 at 16:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We are talking about partitions.
https://en.wikipedia.org/wiki/Partition_(number_theory)
In your case, they are
9
8-1
7-2
7-1-1
6-3
6-2-1
6-1-1-1
5-4
5-3-1
5-2-2
5-2-1-1
5-1-1-1-1
4-4-1
4-3-2
4-3-1-1
4-2-2-1
4-2-1-1-1
4-1-1-1-1-1
3-3-3
3-3-2-1
3-3-1-1-1
3-2-2-2
3-2-2-1-1
3-2-1-1-1-1
3-1-1-1-1-1-1
2-2-2-2-1
2-2-2-1-1-1
2-2-1-1-1-1-1
2-1-1-1-1-1-1-1
1-1-1-1-1-1-1-1-1
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
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add a comment |
$begingroup$
Assuming the boxes are distinguishable, we can put up to 8 bars between 9 objects, each bar separating two boxes. There are 2^8 ways to set the bars, 2 meaning a bar is either there or not. So the answer is 256.
Assuming the boxes are indistinguishable, the question becomes sum of number seperation. There are 1 one to use only 1 box, 8 ways to use 2 boxes, 7 ways to use 3 boxes (711,621,531,522,441,432,333), etc.
answered Jan 14 at 16:43
The PianistThe Pianist
83
$endgroup$
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We are talking about partitions.
https://en.wikipedia.org/wiki/Partition_(number_theory)
In your case, they are
9
8-1
7-2
7-1-1
6-3
6-2-1
6-1-1-1
5-4
5-3-1
5-2-2
5-2-1-1
5-1-1-1-1
4-4-1
4-3-2
4-3-1-1
4-2-2-1
4-2-1-1-1
4-1-1-1-1-1
3-3-3
3-3-2-1
3-3-1-1-1
3-2-2-2
3-2-2-1-1
3-2-1-1-1-1
3-1-1-1-1-1-1
2-2-2-2-1
2-2-2-1-1-1
2-2-1-1-1-1-1
2-1-1-1-1-1-1-1
1-1-1-1-1-1-1-1-1
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
$endgroup$
add a comment |
$begingroup$
We are talking about partitions.
https://en.wikipedia.org/wiki/Partition_(number_theory)
In your case, they are
9
8-1
7-2
7-1-1
6-3
6-2-1
6-1-1-1
5-4
5-3-1
5-2-2
5-2-1-1
5-1-1-1-1
4-4-1
4-3-2
4-3-1-1
4-2-2-1
4-2-1-1-1
4-1-1-1-1-1
3-3-3
3-3-2-1
3-3-1-1-1
3-2-2-2
3-2-2-1-1
3-2-1-1-1-1
3-1-1-1-1-1-1
2-2-2-2-1
2-2-2-1-1-1
2-2-1-1-1-1-1
2-1-1-1-1-1-1-1
1-1-1-1-1-1-1-1-1
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
$endgroup$
add a comment |
$begingroup$
We are talking about partitions.
https://en.wikipedia.org/wiki/Partition_(number_theory)
In your case, they are
9
8-1
7-2
7-1-1
6-3
6-2-1
6-1-1-1
5-4
5-3-1
5-2-2
5-2-1-1
5-1-1-1-1
4-4-1
4-3-2
4-3-1-1
4-2-2-1
4-2-1-1-1
4-1-1-1-1-1
3-3-3
3-3-2-1
3-3-1-1-1
3-2-2-2
3-2-2-1-1
3-2-1-1-1-1
3-1-1-1-1-1-1
2-2-2-2-1
2-2-2-1-1-1
2-2-1-1-1-1-1
2-1-1-1-1-1-1-1
1-1-1-1-1-1-1-1-1
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
$endgroup$
We are talking about partitions.
https://en.wikipedia.org/wiki/Partition_(number_theory)
In your case, they are
9
8-1
7-2
7-1-1
6-3
6-2-1
6-1-1-1
5-4
5-3-1
5-2-2
5-2-1-1
5-1-1-1-1
4-4-1
4-3-2
4-3-1-1
4-2-2-1
4-2-1-1-1
4-1-1-1-1-1
3-3-3
3-3-2-1
3-3-1-1-1
3-2-2-2
3-2-2-1-1
3-2-1-1-1-1
3-1-1-1-1-1-1
2-2-2-2-1
2-2-2-1-1-1
2-2-1-1-1-1-1
2-1-1-1-1-1-1-1
1-1-1-1-1-1-1-1-1
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
answered Jan 14 at 18:13
cgiovanardicgiovanardi
766411
766411
add a comment |
add a comment |
$begingroup$
Assuming the boxes are distinguishable, we can put up to 8 bars between 9 objects, each bar separating two boxes. There are 2^8 ways to set the bars, 2 meaning a bar is either there or not. So the answer is 256.
Assuming the boxes are indistinguishable, the question becomes sum of number seperation. There are 1 one to use only 1 box, 8 ways to use 2 boxes, 7 ways to use 3 boxes (711,621,531,522,441,432,333), etc.
answered Jan 14 at 16:43
The PianistThe Pianist
83
$endgroup$
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
add a comment |
$begingroup$
Assuming the boxes are distinguishable, we can put up to 8 bars between 9 objects, each bar separating two boxes. There are 2^8 ways to set the bars, 2 meaning a bar is either there or not. So the answer is 256.
Assuming the boxes are indistinguishable, the question becomes sum of number seperation. There are 1 one to use only 1 box, 8 ways to use 2 boxes, 7 ways to use 3 boxes (711,621,531,522,441,432,333), etc.
answered Jan 14 at 16:43
The PianistThe Pianist
83
$endgroup$
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
add a comment |
$begingroup$
Assuming the boxes are distinguishable, we can put up to 8 bars between 9 objects, each bar separating two boxes. There are 2^8 ways to set the bars, 2 meaning a bar is either there or not. So the answer is 256.
Assuming the boxes are indistinguishable, the question becomes sum of number seperation. There are 1 one to use only 1 box, 8 ways to use 2 boxes, 7 ways to use 3 boxes (711,621,531,522,441,432,333), etc.
answered Jan 14 at 16:43
The PianistThe Pianist
83
$endgroup$
Assuming the boxes are distinguishable, we can put up to 8 bars between 9 objects, each bar separating two boxes. There are 2^8 ways to set the bars, 2 meaning a bar is either there or not. So the answer is 256.
Assuming the boxes are indistinguishable, the question becomes sum of number seperation. There are 1 one to use only 1 box, 8 ways to use 2 boxes, 7 ways to use 3 boxes (711,621,531,522,441,432,333), etc.
answered Jan 14 at 16:43
The PianistThe Pianist
83
answered Jan 14 at 16:43
The PianistThe Pianist
83
answered Jan 14 at 16:43
The PianistThe Pianist
83
answered Jan 14 at 16:43
The PianistThe Pianist
83
83
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
add a comment |
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong?
$endgroup$
– Stratocarter
Jan 14 at 16:49
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
$begingroup$
Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea.
$endgroup$
– The Pianist
Jan 14 at 17:08
add a comment |
1
$begingroup$
Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second?
$endgroup$
– saulspatz
Jan 14 at 16:29
$begingroup$
The boxes are not distinguishable.
$endgroup$
– Stratocarter
Jan 14 at 16:33
$begingroup$
Then it's partitions, as you said. (I haven't checked that the answer is $30$.)
$endgroup$
– saulspatz
Jan 14 at 16:34
$begingroup$
Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second.
$endgroup$
– Stratocarter
Jan 14 at 16:34