No. Of ways of folding a paper strip Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Number of ways to divide a stick of integer length $N$Folding a strip of paperNumber of ways of partitioning $a+b$ objects into $k $ partitions such that every partition has at least one objectFold, Gather, CutNumber of ways to divide a stick of integer length $N$, take 2A long strip of paperHow many ways we can find out the strip of N spins contains m “Parallel Pair” out of which m1 of them are “Up Parallel Pair”?Find the number of ways in which the number 30 can be partitioned into three unequal partsWhy my process is wrong:-How many ways are there to choose $5$ questions from three sets of $4$, with at least one from each set?combinatorics problems on number of ways a student can answer a a paper of two sections .
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No. Of ways of folding a paper strip
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Number of ways to divide a stick of integer length $N$Folding a strip of paperNumber of ways of partitioning $a+b$ objects into $k $ partitions such that every partition has at least one objectFold, Gather, CutNumber of ways to divide a stick of integer length $N$, take 2A long strip of paperHow many ways we can find out the strip of N spins contains m “Parallel Pair” out of which m1 of them are “Up Parallel Pair”?Find the number of ways in which the number 30 can be partitioned into three unequal partsWhy my process is wrong:-How many ways are there to choose $5$ questions from three sets of $4$, with at least one from each set?combinatorics problems on number of ways a student can answer a a paper of two sections .
$begingroup$
You have given a strip which is divided into n+1 identical parts by n folds on the strip. Now find the no. Of ways in which you can fold the whole strip into a single identical part..!
combinatorics integer-partitions
asked Mar 31 at 18:50
user647127user647127
161
$endgroup$
add a comment |
$begingroup$
You have given a strip which is divided into n+1 identical parts by n folds on the strip. Now find the no. Of ways in which you can fold the whole strip into a single identical part..!
combinatorics integer-partitions
asked Mar 31 at 18:50
user647127user647127
161
$endgroup$
$begingroup$
This was a problem in Martin Gardner's column in Scientific America decades ago.
$endgroup$
– John Wayland Bales
Mar 31 at 18:56
$begingroup$
Here is a discussion of folding a strip of stamps.
$endgroup$
– John Wayland Bales
Mar 31 at 18:59
$begingroup$
But he doesn't provide answer to my question about
$endgroup$
– user647127
Mar 31 at 19:20
$begingroup$
See OEIS A000136
$endgroup$
– Ross Millikan
Mar 31 at 21:04
$begingroup$
But he does make the claim "There seems not to be a closed-form formula to compute the counts of labeled stamp foldings, though many have tried."
$endgroup$
– John Wayland Bales
Mar 31 at 23:23
add a comment |
$begingroup$
You have given a strip which is divided into n+1 identical parts by n folds on the strip. Now find the no. Of ways in which you can fold the whole strip into a single identical part..!
combinatorics integer-partitions
asked Mar 31 at 18:50
user647127user647127
161
$endgroup$
You have given a strip which is divided into n+1 identical parts by n folds on the strip. Now find the no. Of ways in which you can fold the whole strip into a single identical part..!
combinatorics integer-partitions
combinatorics integer-partitions
asked Mar 31 at 18:50
user647127user647127
161
asked Mar 31 at 18:50
user647127user647127
161
asked Mar 31 at 18:50
user647127user647127
161
asked Mar 31 at 18:50
user647127user647127
161
asked Mar 31 at 18:50
user647127user647127
161
161
$begingroup$
This was a problem in Martin Gardner's column in Scientific America decades ago.
$endgroup$
– John Wayland Bales
Mar 31 at 18:56
$begingroup$
Here is a discussion of folding a strip of stamps.
$endgroup$
– John Wayland Bales
Mar 31 at 18:59
$begingroup$
But he doesn't provide answer to my question about
$endgroup$
– user647127
Mar 31 at 19:20
$begingroup$
See OEIS A000136
$endgroup$
– Ross Millikan
Mar 31 at 21:04
$begingroup$
But he does make the claim "There seems not to be a closed-form formula to compute the counts of labeled stamp foldings, though many have tried."
$endgroup$
– John Wayland Bales
Mar 31 at 23:23
add a comment |
$begingroup$
This was a problem in Martin Gardner's column in Scientific America decades ago.
$endgroup$
– John Wayland Bales
Mar 31 at 18:56
$begingroup$
Here is a discussion of folding a strip of stamps.
$endgroup$
– John Wayland Bales
Mar 31 at 18:59
$begingroup$
But he doesn't provide answer to my question about
$endgroup$
– user647127
Mar 31 at 19:20
$begingroup$
See OEIS A000136
$endgroup$
– Ross Millikan
Mar 31 at 21:04
$begingroup$
But he does make the claim "There seems not to be a closed-form formula to compute the counts of labeled stamp foldings, though many have tried."
$endgroup$
– John Wayland Bales
Mar 31 at 23:23
$begingroup$
This was a problem in Martin Gardner's column in Scientific America decades ago.
$endgroup$
– John Wayland Bales
Mar 31 at 18:56
$begingroup$
This was a problem in Martin Gardner's column in Scientific America decades ago.
$endgroup$
– John Wayland Bales
Mar 31 at 18:56
$begingroup$
Here is a discussion of folding a strip of stamps.
$endgroup$
– John Wayland Bales
Mar 31 at 18:59
$begingroup$
Here is a discussion of folding a strip of stamps.
$endgroup$
– John Wayland Bales
Mar 31 at 18:59
$begingroup$
But he doesn't provide answer to my question about
$endgroup$
– user647127
Mar 31 at 19:20
$begingroup$
But he doesn't provide answer to my question about
$endgroup$
– user647127
Mar 31 at 19:20
$begingroup$
See OEIS A000136
$endgroup$
– Ross Millikan
Mar 31 at 21:04
$begingroup$
See OEIS A000136
$endgroup$
– Ross Millikan
Mar 31 at 21:04
$begingroup$
But he does make the claim "There seems not to be a closed-form formula to compute the counts of labeled stamp foldings, though many have tried."
$endgroup$
– John Wayland Bales
Mar 31 at 23:23
$begingroup$
But he does make the claim "There seems not to be a closed-form formula to compute the counts of labeled stamp foldings, though many have tried."
$endgroup$
– John Wayland Bales
Mar 31 at 23:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With the restriction imposed by $n$ folds producing $n+1$ identical parts (folding the strip in half and in half again results in $4$ identical parts but there are actually $3$ folds in the strip's single thickness), then the number of ways to fold the strip into identical parts is $2^n$. Conceptually, each fold can go in one of two directions so the $2$ possibilities of the first fold is followed by $2$ possibilities for the next fold etc. Each resulting configuration of folds has a similar configuration with surface A flipped with surface B. There are practical limitations of actually doing this for some folding configurations as n gets large.
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
$endgroup$
1
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
With the restriction imposed by $n$ folds producing $n+1$ identical parts (folding the strip in half and in half again results in $4$ identical parts but there are actually $3$ folds in the strip's single thickness), then the number of ways to fold the strip into identical parts is $2^n$. Conceptually, each fold can go in one of two directions so the $2$ possibilities of the first fold is followed by $2$ possibilities for the next fold etc. Each resulting configuration of folds has a similar configuration with surface A flipped with surface B. There are practical limitations of actually doing this for some folding configurations as n gets large.
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
$endgroup$
1
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
add a comment |
$begingroup$
With the restriction imposed by $n$ folds producing $n+1$ identical parts (folding the strip in half and in half again results in $4$ identical parts but there are actually $3$ folds in the strip's single thickness), then the number of ways to fold the strip into identical parts is $2^n$. Conceptually, each fold can go in one of two directions so the $2$ possibilities of the first fold is followed by $2$ possibilities for the next fold etc. Each resulting configuration of folds has a similar configuration with surface A flipped with surface B. There are practical limitations of actually doing this for some folding configurations as n gets large.
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
$endgroup$
1
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
add a comment |
$begingroup$
With the restriction imposed by $n$ folds producing $n+1$ identical parts (folding the strip in half and in half again results in $4$ identical parts but there are actually $3$ folds in the strip's single thickness), then the number of ways to fold the strip into identical parts is $2^n$. Conceptually, each fold can go in one of two directions so the $2$ possibilities of the first fold is followed by $2$ possibilities for the next fold etc. Each resulting configuration of folds has a similar configuration with surface A flipped with surface B. There are practical limitations of actually doing this for some folding configurations as n gets large.
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
$endgroup$
With the restriction imposed by $n$ folds producing $n+1$ identical parts (folding the strip in half and in half again results in $4$ identical parts but there are actually $3$ folds in the strip's single thickness), then the number of ways to fold the strip into identical parts is $2^n$. Conceptually, each fold can go in one of two directions so the $2$ possibilities of the first fold is followed by $2$ possibilities for the next fold etc. Each resulting configuration of folds has a similar configuration with surface A flipped with surface B. There are practical limitations of actually doing this for some folding configurations as n gets large.
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
answered Mar 31 at 20:40
Phil HPhil H
4,3102312
4,3102312
1
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
add a comment |
1
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
1
1
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
But for $n=3$ you can have mountain folds between $1$ and $2$ and between $2$ and $3$, then a valley fold between $3$ and $4$. The squares can then be stacked $2134$ or $2341$ I believe these should be counted as distinct.
$endgroup$
– Ross Millikan
Mar 31 at 21:01
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
$begingroup$
@Ross Millikan I see what you are saying, for identical folds the strip can be stacked in different ways. The way I interpreted the question was simply a fold configuration. For n=3, this would result in 12 configurations. I'll look to see if there is a pattern for increasing n. Thanks
$endgroup$
– Phil H
Apr 1 at 14:44
add a comment |
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$begingroup$
This was a problem in Martin Gardner's column in Scientific America decades ago.
$endgroup$
– John Wayland Bales
Mar 31 at 18:56
$begingroup$
Here is a discussion of folding a strip of stamps.
$endgroup$
– John Wayland Bales
Mar 31 at 18:59
$begingroup$
But he doesn't provide answer to my question about
$endgroup$
– user647127
Mar 31 at 19:20
$begingroup$
See OEIS A000136
$endgroup$
– Ross Millikan
Mar 31 at 21:04
$begingroup$
But he does make the claim "There seems not to be a closed-form formula to compute the counts of labeled stamp foldings, though many have tried."
$endgroup$
– John Wayland Bales
Mar 31 at 23:23