Combinations of red and black balls
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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 Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
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|
show 1 more comment
$begingroup$
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 Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
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$begingroup$
I'm pretty sure there is no closed form.
$endgroup$
– Don Thousand
Jan 4 at 17:09
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@DonThousand No closed form as in ?
$endgroup$
– Gaurav Gupta
Jan 4 at 17:19
$begingroup$
No general formula
$endgroup$
– Don Thousand
Jan 4 at 17:20
$begingroup$
@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 :(
$endgroup$
– Gaurav Gupta
Jan 4 at 17:23
$begingroup$
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
$endgroup$
– Gaurav Gupta
Jan 4 at 17:43
|
show 1 more comment
$begingroup$
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 Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
$endgroup$
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 Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
asked Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
edited Jan 4 at 17:44
Gaurav Gupta
asked Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
asked Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
asked Jan 4 at 17:01
Gaurav GuptaGaurav Gupta
212
212
$begingroup$
I'm pretty sure there is no closed form.
$endgroup$
– Don Thousand
Jan 4 at 17:09
$begingroup$
@DonThousand No closed form as in ?
$endgroup$
– Gaurav Gupta
Jan 4 at 17:19
$begingroup$
No general formula
$endgroup$
– Don Thousand
Jan 4 at 17:20
$begingroup$
@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 :(
$endgroup$
– Gaurav Gupta
Jan 4 at 17:23
$begingroup$
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
$endgroup$
– Gaurav Gupta
Jan 4 at 17:43
|
show 1 more comment
$begingroup$
I'm pretty sure there is no closed form.
$endgroup$
– Don Thousand
Jan 4 at 17:09
$begingroup$
@DonThousand No closed form as in ?
$endgroup$
– Gaurav Gupta
Jan 4 at 17:19
$begingroup$
No general formula
$endgroup$
– Don Thousand
Jan 4 at 17:20
$begingroup$
@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 :(
$endgroup$
– Gaurav Gupta
Jan 4 at 17:23
$begingroup$
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
$endgroup$
– Gaurav Gupta
Jan 4 at 17:43
$begingroup$
I'm pretty sure there is no closed form.
$endgroup$
– Don Thousand
Jan 4 at 17:09
$begingroup$
I'm pretty sure there is no closed form.
$endgroup$
– Don Thousand
Jan 4 at 17:09
$begingroup$
@DonThousand No closed form as in ?
$endgroup$
– Gaurav Gupta
Jan 4 at 17:19
$begingroup$
@DonThousand No closed form as in ?
$endgroup$
– Gaurav Gupta
Jan 4 at 17:19
$begingroup$
No general formula
$endgroup$
– Don Thousand
Jan 4 at 17:20
$begingroup$
No general formula
$endgroup$
– Don Thousand
Jan 4 at 17:20
$begingroup$
@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 :(
$endgroup$
– Gaurav Gupta
Jan 4 at 17:23
$begingroup$
@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 :(
$endgroup$
– Gaurav Gupta
Jan 4 at 17:23
$begingroup$
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
$endgroup$
– Gaurav Gupta
Jan 4 at 17:43
$begingroup$
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
$endgroup$
– Gaurav Gupta
Jan 4 at 17:43
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$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 Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
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add a comment |
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1 Answer
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$begingroup$
$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 Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
$endgroup$
add a comment |
$begingroup$
$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 Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
$endgroup$
add a comment |
$begingroup$
$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 Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
$endgroup$
$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 Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
answered Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
answered Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
answered Jan 5 at 11:40
Zachary HunterZachary Hunter
1,017313
1,017313
add a comment |
add a comment |
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$begingroup$
I'm pretty sure there is no closed form.
$endgroup$
– Don Thousand
Jan 4 at 17:09
$begingroup$
@DonThousand No closed form as in ?
$endgroup$
– Gaurav Gupta
Jan 4 at 17:19
$begingroup$
No general formula
$endgroup$
– Don Thousand
Jan 4 at 17:20
$begingroup$
@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 :(
$endgroup$
– Gaurav Gupta
Jan 4 at 17:23
$begingroup$
@DonThousand Looks like there exist a dynamic programming solution to this to find it.
$endgroup$
– Gaurav Gupta
Jan 4 at 17:43