Combinatorics: $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together.
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$N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?
This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!
It is not necessary that women and men alternate, ACaBcb
combinatorics
edited Nov 18 at 7:56
Robert Z
90.7k1057128
asked Nov 18 at 7:11
kevin
1056
add a comment |
up vote
1
down vote
favorite
$N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?
This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!
It is not necessary that women and men alternate, ACaBcb
combinatorics
edited Nov 18 at 7:56
Robert Z
90.7k1057128
asked Nov 18 at 7:11
kevin
1056
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?
This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!
It is not necessary that women and men alternate, ACaBcb
combinatorics
edited Nov 18 at 7:56
Robert Z
90.7k1057128
asked Nov 18 at 7:11
kevin
1056
$N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?
This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!
It is not necessary that women and men alternate, ACaBcb
combinatorics
combinatorics
edited Nov 18 at 7:56
Robert Z
90.7k1057128
asked Nov 18 at 7:11
kevin
1056
edited Nov 18 at 7:56
Robert Z
90.7k1057128
asked Nov 18 at 7:11
kevin
1056
edited Nov 18 at 7:56
Robert Z
90.7k1057128
edited Nov 18 at 7:56
Robert Z
90.7k1057128
edited Nov 18 at 7:56
Robert Z
90.7k1057128
90.7k1057128
asked Nov 18 at 7:11
kevin
1056
asked Nov 18 at 7:11
kevin
1056
asked Nov 18 at 7:11
kevin
1056
1056
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint. By the inclusion-exclusion principle we have
$$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
Can you take it from here?
answered Nov 18 at 7:31
Robert Z
90.7k1057128
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
1
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
1
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint. By the inclusion-exclusion principle we have
$$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
Can you take it from here?
answered Nov 18 at 7:31
Robert Z
90.7k1057128
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
1
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
1
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
|
show 2 more comments
up vote
1
down vote
accepted
Hint. By the inclusion-exclusion principle we have
$$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
Can you take it from here?
answered Nov 18 at 7:31
Robert Z
90.7k1057128
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
1
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
1
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
|
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint. By the inclusion-exclusion principle we have
$$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
Can you take it from here?
answered Nov 18 at 7:31
Robert Z
90.7k1057128
Hint. By the inclusion-exclusion principle we have
$$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
Can you take it from here?
answered Nov 18 at 7:31
Robert Z
90.7k1057128
edited Nov 18 at 8:47
answered Nov 18 at 7:31
Robert Z
90.7k1057128
answered Nov 18 at 7:31
Robert Z
90.7k1057128
answered Nov 18 at 7:31
Robert Z
90.7k1057128
90.7k1057128
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
1
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
1
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
|
show 2 more comments
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
1
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
1
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
Thanks! Alternate till M couples right?
– kevin
Nov 18 at 7:38
1
1
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
@kevin Yes, that's correct.
– Robert Z
Nov 18 at 7:39
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"？ Suppose that's true, why alternate with additional terms? Thanks!
– kevin
Nov 18 at 8:29
1
1
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
– Robert Z
Nov 18 at 8:44
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
– kevin
Nov 18 at 9:22
|
show 2 more comments
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