Combinations of red and black balls
Given $N$ Identical Red balls and $M$ Identical Black balls, in how many ways we can arrange them such that not more than $K$ adjacent balls are of same color.
Example : For $1$ Red ball and $1$ black ball, with $K=1$, there are $2$ ways $[RB,BR]$
Can there be a general formula for given $N$,$M$ and $K$ ?
I have read about Dutch flag problem to find number of ways to find such that no adjacent balls are of same color. I am bit stuck on how to find for at max K balls.
combinatorics number-theory algorithms dynamic-programming
asked yesterday
Gaurav Gupta
212
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Given $N$ Identical Red balls and $M$ Identical Black balls, in how many ways we can arrange them such that not more than $K$ adjacent balls are of same color.
Example : For $1$ Red ball and $1$ black ball, with $K=1$, there are $2$ ways $[RB,BR]$
Can there be a general formula for given $N$,$M$ and $K$ ?
I have read about Dutch flag problem to find number of ways to find such that no adjacent balls are of same color. I am bit stuck on how to find for at max K balls.
combinatorics number-theory algorithms dynamic-programming
asked yesterday
Gaurav Gupta
212
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm pretty sure there is no closed form.
– Don Thousand
yesterday
@DonThousand No closed form as in ?
– Gaurav Gupta
yesterday
No general formula
– Don Thousand
yesterday
@DonThousand Ah I see. I was thinking of something like, if 1 adjacent ball can be of same color then how many ways + if 2 adjacent balls can be of same color then how manys and so on upto K. Wasn't able to have a general solution :(
– Gaurav Gupta
23 hours ago
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
– Gaurav Gupta
23 hours ago
|
show 1 more comment
Given $N$ Identical Red balls and $M$ Identical Black balls, in how many ways we can arrange them such that not more than $K$ adjacent balls are of same color.
Example : For $1$ Red ball and $1$ black ball, with $K=1$, there are $2$ ways $[RB,BR]$
Can there be a general formula for given $N$,$M$ and $K$ ?
I have read about Dutch flag problem to find number of ways to find such that no adjacent balls are of same color. I am bit stuck on how to find for at max K balls.
combinatorics number-theory algorithms dynamic-programming
asked yesterday
Gaurav Gupta
212
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given $N$ Identical Red balls and $M$ Identical Black balls, in how many ways we can arrange them such that not more than $K$ adjacent balls are of same color.
Example : For $1$ Red ball and $1$ black ball, with $K=1$, there are $2$ ways $[RB,BR]$
Can there be a general formula for given $N$,$M$ and $K$ ?
I have read about Dutch flag problem to find number of ways to find such that no adjacent balls are of same color. I am bit stuck on how to find for at max K balls.
combinatorics number-theory algorithms dynamic-programming
combinatorics number-theory algorithms dynamic-programming
asked yesterday
Gaurav Gupta
212
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Gaurav Gupta
212
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 hours ago
asked yesterday
Gaurav Gupta
212
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Gaurav Gupta
212
asked yesterday
Gaurav Gupta
212
212
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gaurav Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm pretty sure there is no closed form.
– Don Thousand
yesterday
@DonThousand No closed form as in ?
– Gaurav Gupta
yesterday
No general formula
– Don Thousand
yesterday
@DonThousand Ah I see. I was thinking of something like, if 1 adjacent ball can be of same color then how many ways + if 2 adjacent balls can be of same color then how manys and so on upto K. Wasn't able to have a general solution :(
– Gaurav Gupta
23 hours ago
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
– Gaurav Gupta
23 hours ago
|
show 1 more comment
I'm pretty sure there is no closed form.
– Don Thousand
yesterday
@DonThousand No closed form as in ?
– Gaurav Gupta
yesterday
No general formula
– Don Thousand
yesterday
@DonThousand Ah I see. I was thinking of something like, if 1 adjacent ball can be of same color then how many ways + if 2 adjacent balls can be of same color then how manys and so on upto K. Wasn't able to have a general solution :(
– Gaurav Gupta
23 hours ago
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
– Gaurav Gupta
23 hours ago
I'm pretty sure there is no closed form.
– Don Thousand
yesterday
I'm pretty sure there is no closed form.
– Don Thousand
yesterday
@DonThousand No closed form as in ?
– Gaurav Gupta
yesterday
@DonThousand No closed form as in ?
– Gaurav Gupta
yesterday
No general formula
– Don Thousand
yesterday
No general formula
– Don Thousand
yesterday
@DonThousand Ah I see. I was thinking of something like, if 1 adjacent ball can be of same color then how many ways + if 2 adjacent balls can be of same color then how manys and so on upto K. Wasn't able to have a general solution :(
– Gaurav Gupta
23 hours ago
@DonThousand Ah I see. I was thinking of something like, if 1 adjacent ball can be of same color then how many ways + if 2 adjacent balls can be of same color then how manys and so on upto K. Wasn't able to have a general solution :(
– Gaurav Gupta
23 hours ago
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
– Gaurav Gupta
23 hours ago
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
– Gaurav Gupta
23 hours ago
|
show 1 more comment
1 Answer
1
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$f(n,k) = $ number of sequences with $n$ balls and exactly $k$ repeats, which is exactly this:
$f(n,k) = 2 binom{n-1}{k}$
Essentially, there are two sequences with 0 repeats and $n-k$ length. Given a string with no repeats, we choose how many extra balls are added into each spot, which is equivalent to multiplying by the multichoose $left( binom{n-k}{k} right) = binom{n-1}{k}$.
I guess you’d want to sum this function for all k less than K, but that’s the nicest idea I could come up with. Apologies for my sloppy phrasing, I can clarify where needed.
answered 5 hours ago
Zachary Hunter
5069
add a comment |
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1 Answer
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$f(n,k) = $ number of sequences with $n$ balls and exactly $k$ repeats, which is exactly this:
$f(n,k) = 2 binom{n-1}{k}$
Essentially, there are two sequences with 0 repeats and $n-k$ length. Given a string with no repeats, we choose how many extra balls are added into each spot, which is equivalent to multiplying by the multichoose $left( binom{n-k}{k} right) = binom{n-1}{k}$.
I guess you’d want to sum this function for all k less than K, but that’s the nicest idea I could come up with. Apologies for my sloppy phrasing, I can clarify where needed.
answered 5 hours ago
Zachary Hunter
5069
add a comment |
$f(n,k) = $ number of sequences with $n$ balls and exactly $k$ repeats, which is exactly this:
$f(n,k) = 2 binom{n-1}{k}$
Essentially, there are two sequences with 0 repeats and $n-k$ length. Given a string with no repeats, we choose how many extra balls are added into each spot, which is equivalent to multiplying by the multichoose $left( binom{n-k}{k} right) = binom{n-1}{k}$.
I guess you’d want to sum this function for all k less than K, but that’s the nicest idea I could come up with. Apologies for my sloppy phrasing, I can clarify where needed.
answered 5 hours ago
Zachary Hunter
5069
add a comment |
$f(n,k) = $ number of sequences with $n$ balls and exactly $k$ repeats, which is exactly this:
$f(n,k) = 2 binom{n-1}{k}$
Essentially, there are two sequences with 0 repeats and $n-k$ length. Given a string with no repeats, we choose how many extra balls are added into each spot, which is equivalent to multiplying by the multichoose $left( binom{n-k}{k} right) = binom{n-1}{k}$.
I guess you’d want to sum this function for all k less than K, but that’s the nicest idea I could come up with. Apologies for my sloppy phrasing, I can clarify where needed.
answered 5 hours ago
Zachary Hunter
5069
$f(n,k) = $ number of sequences with $n$ balls and exactly $k$ repeats, which is exactly this:
$f(n,k) = 2 binom{n-1}{k}$
Essentially, there are two sequences with 0 repeats and $n-k$ length. Given a string with no repeats, we choose how many extra balls are added into each spot, which is equivalent to multiplying by the multichoose $left( binom{n-k}{k} right) = binom{n-1}{k}$.
I guess you’d want to sum this function for all k less than K, but that’s the nicest idea I could come up with. Apologies for my sloppy phrasing, I can clarify where needed.
answered 5 hours ago
Zachary Hunter
5069
answered 5 hours ago
Zachary Hunter
5069
answered 5 hours ago
Zachary Hunter
5069
answered 5 hours ago
Zachary Hunter
5069
5069
add a comment |
add a comment |
Gaurav Gupta is a new contributor. Be nice, and check out our Code of Conduct.
Gaurav Gupta is a new contributor. Be nice, and check out our Code of Conduct.
Gaurav Gupta is a new contributor. Be nice, and check out our Code of Conduct.
Gaurav Gupta is a new contributor. Be nice, and check out our Code of Conduct.
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I'm pretty sure there is no closed form.
– Don Thousand
yesterday
@DonThousand No closed form as in ?
– Gaurav Gupta
yesterday
No general formula
– Don Thousand
yesterday
@DonThousand Ah I see. I was thinking of something like, if 1 adjacent ball can be of same color then how many ways + if 2 adjacent balls can be of same color then how manys and so on upto K. Wasn't able to have a general solution :(
– Gaurav Gupta
23 hours ago
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
– Gaurav Gupta
23 hours ago