In how many ways can we put $k$ sticks between $n$ circles?
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Let’s suppose we have $k$ sticks and we want to put them between $n$ circles where $n$ and $k$ are $2$ natural numbers.
In how many ways can we put the sticks between the circles?
Please note that the conditions are:
(I’m using an example with $2$ sticks and $5$ circles:)
$1 cdot$The $k$ sticks are identical.For example if you put two sticks between the first and the second circle, you can’t make a new combination by switching them.
$2 cdot$You can’t change the position of the circles.Note that the circles are not identical so if you put the two sticks between the first and the second circle , you can make a new combination by putting the two sticks between the fourth and the fifth circles.
$3 cdot$You can put the sticks wherever you want.(They can also be put next to each other with no circles between them. Also, you can put them outside the circles so that means that there can be only one circle to the right/left of a stick.
Here’s what I’ve tried:
I thought since there are $k$ sticks, that means that the sticks make $k+1$ sections, where there could be $0$ to $n$ circles in each of the sections.
So:
If we find how many $k+1$ non-negative integers there are such that their sum is $n$ , we’ll find the answer.
combinatorics
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
asked Aug 21 at 7:48
Borna Ahmadzade
297
 |Â
show 1 more comment
up vote
2
down vote
favorite
Let’s suppose we have $k$ sticks and we want to put them between $n$ circles where $n$ and $k$ are $2$ natural numbers.
In how many ways can we put the sticks between the circles?
Please note that the conditions are:
(I’m using an example with $2$ sticks and $5$ circles:)
$1 cdot$The $k$ sticks are identical.For example if you put two sticks between the first and the second circle, you can’t make a new combination by switching them.
$2 cdot$You can’t change the position of the circles.Note that the circles are not identical so if you put the two sticks between the first and the second circle , you can make a new combination by putting the two sticks between the fourth and the fifth circles.
$3 cdot$You can put the sticks wherever you want.(They can also be put next to each other with no circles between them. Also, you can put them outside the circles so that means that there can be only one circle to the right/left of a stick.
Here’s what I’ve tried:
I thought since there are $k$ sticks, that means that the sticks make $k+1$ sections, where there could be $0$ to $n$ circles in each of the sections.
So:
If we find how many $k+1$ non-negative integers there are such that their sum is $n$ , we’ll find the answer.
combinatorics
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
asked Aug 21 at 7:48
Borna Ahmadzade
297
1
You should take a look at stars and bars.
– PM 2Ring
Aug 21 at 7:55
1
Hint: You have $n+k$ slots to fill, and you want to choose $k$ slots for the sticks, leaving $n$ slots for the circles.
– PM 2Ring
Aug 21 at 8:22
@PM2RingYou should add it as an answer. I did but noticed you already said it in comments. So, deleted.
– SinTan1729
Aug 21 at 8:27
@PM2Ring I have a question : Why n+k slots? There are n circles placed in a row and n-1 slots between them but you could also place the sticks not between two circles but to the left of the very left circle and to the right of the very right circle which would be n + 1.Can yoy explain please?
– Borna Ahmadzade
Aug 21 at 9:59
1
There are $n$ circles and $k$ sticks in a row, so there are a total of $n+k$ objects in a row. If all the objects were unique, then there would be $(n+k)!$ ways to arrange them in a row.
– PM 2Ring
Aug 21 at 11:03
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let’s suppose we have $k$ sticks and we want to put them between $n$ circles where $n$ and $k$ are $2$ natural numbers.
In how many ways can we put the sticks between the circles?
Please note that the conditions are:
(I’m using an example with $2$ sticks and $5$ circles:)
$1 cdot$The $k$ sticks are identical.For example if you put two sticks between the first and the second circle, you can’t make a new combination by switching them.
$2 cdot$You can’t change the position of the circles.Note that the circles are not identical so if you put the two sticks between the first and the second circle , you can make a new combination by putting the two sticks between the fourth and the fifth circles.
$3 cdot$You can put the sticks wherever you want.(They can also be put next to each other with no circles between them. Also, you can put them outside the circles so that means that there can be only one circle to the right/left of a stick.
Here’s what I’ve tried:
I thought since there are $k$ sticks, that means that the sticks make $k+1$ sections, where there could be $0$ to $n$ circles in each of the sections.
So:
If we find how many $k+1$ non-negative integers there are such that their sum is $n$ , we’ll find the answer.
combinatorics
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
asked Aug 21 at 7:48
Borna Ahmadzade
297
Let’s suppose we have $k$ sticks and we want to put them between $n$ circles where $n$ and $k$ are $2$ natural numbers.
In how many ways can we put the sticks between the circles?
Please note that the conditions are:
(I’m using an example with $2$ sticks and $5$ circles:)
$1 cdot$The $k$ sticks are identical.For example if you put two sticks between the first and the second circle, you can’t make a new combination by switching them.
$2 cdot$You can’t change the position of the circles.Note that the circles are not identical so if you put the two sticks between the first and the second circle , you can make a new combination by putting the two sticks between the fourth and the fifth circles.
$3 cdot$You can put the sticks wherever you want.(They can also be put next to each other with no circles between them. Also, you can put them outside the circles so that means that there can be only one circle to the right/left of a stick.
Here’s what I’ve tried:
I thought since there are $k$ sticks, that means that the sticks make $k+1$ sections, where there could be $0$ to $n$ circles in each of the sections.
So:
If we find how many $k+1$ non-negative integers there are such that their sum is $n$ , we’ll find the answer.
combinatorics
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
asked Aug 21 at 7:48
Borna Ahmadzade
297
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
edited Aug 21 at 8:36
N. F. Taussig
38.8k93153
38.8k93153
asked Aug 21 at 7:48
Borna Ahmadzade
297
asked Aug 21 at 7:48
Borna Ahmadzade
297
asked Aug 21 at 7:48
Borna Ahmadzade
297
297
1
You should take a look at stars and bars.
– PM 2Ring
Aug 21 at 7:55
1
Hint: You have $n+k$ slots to fill, and you want to choose $k$ slots for the sticks, leaving $n$ slots for the circles.
– PM 2Ring
Aug 21 at 8:22
@PM2RingYou should add it as an answer. I did but noticed you already said it in comments. So, deleted.
– SinTan1729
Aug 21 at 8:27
@PM2Ring I have a question : Why n+k slots? There are n circles placed in a row and n-1 slots between them but you could also place the sticks not between two circles but to the left of the very left circle and to the right of the very right circle which would be n + 1.Can yoy explain please?
– Borna Ahmadzade
Aug 21 at 9:59
1
There are $n$ circles and $k$ sticks in a row, so there are a total of $n+k$ objects in a row. If all the objects were unique, then there would be $(n+k)!$ ways to arrange them in a row.
– PM 2Ring
Aug 21 at 11:03
 |Â
show 1 more comment
1
You should take a look at stars and bars.
– PM 2Ring
Aug 21 at 7:55
1
Hint: You have $n+k$ slots to fill, and you want to choose $k$ slots for the sticks, leaving $n$ slots for the circles.
– PM 2Ring
Aug 21 at 8:22
@PM2RingYou should add it as an answer. I did but noticed you already said it in comments. So, deleted.
– SinTan1729
Aug 21 at 8:27
@PM2Ring I have a question : Why n+k slots? There are n circles placed in a row and n-1 slots between them but you could also place the sticks not between two circles but to the left of the very left circle and to the right of the very right circle which would be n + 1.Can yoy explain please?
– Borna Ahmadzade
Aug 21 at 9:59
1
There are $n$ circles and $k$ sticks in a row, so there are a total of $n+k$ objects in a row. If all the objects were unique, then there would be $(n+k)!$ ways to arrange them in a row.
– PM 2Ring
Aug 21 at 11:03
1
1
You should take a look at stars and bars.
– PM 2Ring
Aug 21 at 7:55
You should take a look at stars and bars.
– PM 2Ring
Aug 21 at 7:55
1
1
Hint: You have $n+k$ slots to fill, and you want to choose $k$ slots for the sticks, leaving $n$ slots for the circles.
– PM 2Ring
Aug 21 at 8:22
Hint: You have $n+k$ slots to fill, and you want to choose $k$ slots for the sticks, leaving $n$ slots for the circles.
– PM 2Ring
Aug 21 at 8:22
@PM2RingYou should add it as an answer. I did but noticed you already said it in comments. So, deleted.
– SinTan1729
Aug 21 at 8:27
@PM2RingYou should add it as an answer. I did but noticed you already said it in comments. So, deleted.
– SinTan1729
Aug 21 at 8:27
@PM2Ring I have a question : Why n+k slots? There are n circles placed in a row and n-1 slots between them but you could also place the sticks not between two circles but to the left of the very left circle and to the right of the very right circle which would be n + 1.Can yoy explain please?
– Borna Ahmadzade
Aug 21 at 9:59
@PM2Ring I have a question : Why n+k slots? There are n circles placed in a row and n-1 slots between them but you could also place the sticks not between two circles but to the left of the very left circle and to the right of the very right circle which would be n + 1.Can yoy explain please?
– Borna Ahmadzade
Aug 21 at 9:59
1
1
There are $n$ circles and $k$ sticks in a row, so there are a total of $n+k$ objects in a row. If all the objects were unique, then there would be $(n+k)!$ ways to arrange them in a row.
– PM 2Ring
Aug 21 at 11:03
There are $n$ circles and $k$ sticks in a row, so there are a total of $n+k$ objects in a row. If all the objects were unique, then there would be $(n+k)!$ ways to arrange them in a row.
– PM 2Ring
Aug 21 at 11:03
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
Since nobody has answered , I’m going to answer my question using @PM2Ring explanation:
So first , we assume that the sticks are not identical and we can change the position of the circles.
So there are $(n+k)!$ combinations.
But we know that we can’t change the position of the circles and also the sticks are identical.
So we have to divide $(n+k)!$ by $k!n!$ where $k!$ is the number of permutations of the sticks that we counted but shouldn’t have because they’re identical and where $n!$ is the number of permutations that the circles would have if we were able to change the position of the circles.
So the answer would be:
P(n+k;n,k)
answered Aug 22 at 8:32
Borna Ahmadzade
297
1
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since nobody has answered , I’m going to answer my question using @PM2Ring explanation:
So first , we assume that the sticks are not identical and we can change the position of the circles.
So there are $(n+k)!$ combinations.
But we know that we can’t change the position of the circles and also the sticks are identical.
So we have to divide $(n+k)!$ by $k!n!$ where $k!$ is the number of permutations of the sticks that we counted but shouldn’t have because they’re identical and where $n!$ is the number of permutations that the circles would have if we were able to change the position of the circles.
So the answer would be:
P(n+k;n,k)
answered Aug 22 at 8:32
Borna Ahmadzade
297
1
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
add a comment |Â
up vote
1
down vote
Since nobody has answered , I’m going to answer my question using @PM2Ring explanation:
So first , we assume that the sticks are not identical and we can change the position of the circles.
So there are $(n+k)!$ combinations.
But we know that we can’t change the position of the circles and also the sticks are identical.
So we have to divide $(n+k)!$ by $k!n!$ where $k!$ is the number of permutations of the sticks that we counted but shouldn’t have because they’re identical and where $n!$ is the number of permutations that the circles would have if we were able to change the position of the circles.
So the answer would be:
P(n+k;n,k)
answered Aug 22 at 8:32
Borna Ahmadzade
297
1
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since nobody has answered , I’m going to answer my question using @PM2Ring explanation:
So first , we assume that the sticks are not identical and we can change the position of the circles.
So there are $(n+k)!$ combinations.
But we know that we can’t change the position of the circles and also the sticks are identical.
So we have to divide $(n+k)!$ by $k!n!$ where $k!$ is the number of permutations of the sticks that we counted but shouldn’t have because they’re identical and where $n!$ is the number of permutations that the circles would have if we were able to change the position of the circles.
So the answer would be:
P(n+k;n,k)
answered Aug 22 at 8:32
Borna Ahmadzade
297
Since nobody has answered , I’m going to answer my question using @PM2Ring explanation:
So first , we assume that the sticks are not identical and we can change the position of the circles.
So there are $(n+k)!$ combinations.
But we know that we can’t change the position of the circles and also the sticks are identical.
So we have to divide $(n+k)!$ by $k!n!$ where $k!$ is the number of permutations of the sticks that we counted but shouldn’t have because they’re identical and where $n!$ is the number of permutations that the circles would have if we were able to change the position of the circles.
So the answer would be:
P(n+k;n,k)
answered Aug 22 at 8:32
Borna Ahmadzade
297
answered Aug 22 at 8:32
Borna Ahmadzade
297
answered Aug 22 at 8:32
Borna Ahmadzade
297
answered Aug 22 at 8:32
Borna Ahmadzade
297
297
1
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
add a comment |Â
1
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
1
1
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
Exactly! And so for your example of 5 circles and 2 sticks there are $7times6/2=21$ permutations: 0000011, 0000101, 0000110, 0001001, 0001010, 0001100, 0010001, 0010010, 0010100, 0011000, 0100001, 0100010, 0100100, 0101000, 0110000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000
– PM 2Ring
Aug 22 at 14:55
add a comment |Â
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1
You should take a look at stars and bars.
– PM 2Ring
Aug 21 at 7:55
1
Hint: You have $n+k$ slots to fill, and you want to choose $k$ slots for the sticks, leaving $n$ slots for the circles.
– PM 2Ring
Aug 21 at 8:22
@PM2RingYou should add it as an answer. I did but noticed you already said it in comments. So, deleted.
– SinTan1729
Aug 21 at 8:27
@PM2Ring I have a question : Why n+k slots? There are n circles placed in a row and n-1 slots between them but you could also place the sticks not between two circles but to the left of the very left circle and to the right of the very right circle which would be n + 1.Can yoy explain please?
– Borna Ahmadzade
Aug 21 at 9:59
1
There are $n$ circles and $k$ sticks in a row, so there are a total of $n+k$ objects in a row. If all the objects were unique, then there would be $(n+k)!$ ways to arrange them in a row.
– PM 2Ring
Aug 21 at 11:03