Permutation of boxes with two colors
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Given a row with $n$ black boxes and $n$ white boxes, how many permutations exist where each box can be adjacent to at most one other box of the same color?
In other words, three or more consecutive boxes with the same color are not allowed.
For $n=3$, I count $14$ permutations and for $n=4$, I get $34$ permutations, but what's the general rule?
permutations recreational-mathematics
asked Jan 4 at 4:47
JensJens
3,90021031
$endgroup$
add a comment |
$begingroup$
Given a row with $n$ black boxes and $n$ white boxes, how many permutations exist where each box can be adjacent to at most one other box of the same color?
In other words, three or more consecutive boxes with the same color are not allowed.
For $n=3$, I count $14$ permutations and for $n=4$, I get $34$ permutations, but what's the general rule?
permutations recreational-mathematics
asked Jan 4 at 4:47
JensJens
3,90021031
$endgroup$
$begingroup$
for n= 3, I count 12 permutations.
$endgroup$
– Doug M
Jan 4 at 5:01
$begingroup$
Similar question, generalized for $k$ colors, but without the restriction that there be equal number of each color: painting fence...
$endgroup$
– Daniel Mathias
Jan 4 at 11:53
add a comment |
$begingroup$
Given a row with $n$ black boxes and $n$ white boxes, how many permutations exist where each box can be adjacent to at most one other box of the same color?
In other words, three or more consecutive boxes with the same color are not allowed.
For $n=3$, I count $14$ permutations and for $n=4$, I get $34$ permutations, but what's the general rule?
permutations recreational-mathematics
asked Jan 4 at 4:47
JensJens
3,90021031
$endgroup$
Given a row with $n$ black boxes and $n$ white boxes, how many permutations exist where each box can be adjacent to at most one other box of the same color?
In other words, three or more consecutive boxes with the same color are not allowed.
For $n=3$, I count $14$ permutations and for $n=4$, I get $34$ permutations, but what's the general rule?
permutations recreational-mathematics
permutations recreational-mathematics
asked Jan 4 at 4:47
JensJens
3,90021031
asked Jan 4 at 4:47
JensJens
3,90021031
asked Jan 4 at 4:47
JensJens
3,90021031
asked Jan 4 at 4:47
JensJens
3,90021031
asked Jan 4 at 4:47
JensJens
3,90021031
3,90021031
$begingroup$
for n= 3, I count 12 permutations.
$endgroup$
– Doug M
Jan 4 at 5:01
$begingroup$
Similar question, generalized for $k$ colors, but without the restriction that there be equal number of each color: painting fence...
$endgroup$
– Daniel Mathias
Jan 4 at 11:53
add a comment |
$begingroup$
for n= 3, I count 12 permutations.
$endgroup$
– Doug M
Jan 4 at 5:01
$begingroup$
Similar question, generalized for $k$ colors, but without the restriction that there be equal number of each color: painting fence...
$endgroup$
– Daniel Mathias
Jan 4 at 11:53
$begingroup$
for n= 3, I count 12 permutations.
$endgroup$
– Doug M
Jan 4 at 5:01
$begingroup$
for n= 3, I count 12 permutations.
$endgroup$
– Doug M
Jan 4 at 5:01
$begingroup$
Similar question, generalized for $k$ colors, but without the restriction that there be equal number of each color: painting fence...
$endgroup$
– Daniel Mathias
Jan 4 at 11:53
$begingroup$
Similar question, generalized for $k$ colors, but without the restriction that there be equal number of each color: painting fence...
$endgroup$
– Daniel Mathias
Jan 4 at 11:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is A177790 “Number of paths from (0,0) to (n,n) avoiding 3 or more consecutive east steps and 3 or more consecutive north steps” on OEIS. The closest thing to a formula given there is
$$
a(n) = sum_{i=0}^{lfloor n/2rfloor} 2binom{n-i}{i}^2 + binom{n-i}{i} binom{n-i-1}{i+1} + binom{n-i}{i}binom{n-i+1}{i-1}.
$$
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is A177790 “Number of paths from (0,0) to (n,n) avoiding 3 or more consecutive east steps and 3 or more consecutive north steps” on OEIS. The closest thing to a formula given there is
$$
a(n) = sum_{i=0}^{lfloor n/2rfloor} 2binom{n-i}{i}^2 + binom{n-i}{i} binom{n-i-1}{i+1} + binom{n-i}{i}binom{n-i+1}{i-1}.
$$
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
$endgroup$
add a comment |
$begingroup$
This is A177790 “Number of paths from (0,0) to (n,n) avoiding 3 or more consecutive east steps and 3 or more consecutive north steps” on OEIS. The closest thing to a formula given there is
$$
a(n) = sum_{i=0}^{lfloor n/2rfloor} 2binom{n-i}{i}^2 + binom{n-i}{i} binom{n-i-1}{i+1} + binom{n-i}{i}binom{n-i+1}{i-1}.
$$
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
$endgroup$
add a comment |
$begingroup$
This is A177790 “Number of paths from (0,0) to (n,n) avoiding 3 or more consecutive east steps and 3 or more consecutive north steps” on OEIS. The closest thing to a formula given there is
$$
a(n) = sum_{i=0}^{lfloor n/2rfloor} 2binom{n-i}{i}^2 + binom{n-i}{i} binom{n-i-1}{i+1} + binom{n-i}{i}binom{n-i+1}{i-1}.
$$
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
$endgroup$
This is A177790 “Number of paths from (0,0) to (n,n) avoiding 3 or more consecutive east steps and 3 or more consecutive north steps” on OEIS. The closest thing to a formula given there is
$$
a(n) = sum_{i=0}^{lfloor n/2rfloor} 2binom{n-i}{i}^2 + binom{n-i}{i} binom{n-i-1}{i+1} + binom{n-i}{i}binom{n-i+1}{i-1}.
$$
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
answered Jan 4 at 5:01
Anders KaseorgAnders Kaseorg
79349
79349
add a comment |
add a comment |
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$begingroup$
for n= 3, I count 12 permutations.
$endgroup$
– Doug M
Jan 4 at 5:01
$begingroup$
Similar question, generalized for $k$ colors, but without the restriction that there be equal number of each color: painting fence...
$endgroup$
– Daniel Mathias
Jan 4 at 11:53