Round table pairs combinatorics
$begingroup$
I know this question will sound easy, but i just cant figure this out. Please help me.
How many ways are there for putting 4 pairs to the round table, so that atleast 1 pair is sitting next to each other.
combinatorics
asked Dec 10 '18 at 17:19
AlletherAllether
31
$endgroup$
add a comment |
$begingroup$
I know this question will sound easy, but i just cant figure this out. Please help me.
How many ways are there for putting 4 pairs to the round table, so that atleast 1 pair is sitting next to each other.
combinatorics
asked Dec 10 '18 at 17:19
AlletherAllether
31
$endgroup$
$begingroup$
How many ways can you pick a pair? How many ways can that pair be seated so that they are next to each other? Then how many ways can you seat everyone else?
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:23
$begingroup$
My 'solution' is this: i take 1 pair aside so i will always have atleast 1 pair next to each other, now for 3 pairs its 5! and now i dont know how to include that 1 pair to this.. my result is 5! + 6*2, because there are 6 places where to put that pair and *2 because they can sit like man-woman and woman-man
$endgroup$
– Allether
Dec 10 '18 at 17:27
$begingroup$
The fact that they are seated at a round table suggests to me that two seatings are equivalent if they are the same "up to rotation".
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:56
$begingroup$
One approach is to subtract the number of arrangements in which no pair is sitting next to each other from the total number of arrangements.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:55
add a comment |
$begingroup$
I know this question will sound easy, but i just cant figure this out. Please help me.
How many ways are there for putting 4 pairs to the round table, so that atleast 1 pair is sitting next to each other.
combinatorics
asked Dec 10 '18 at 17:19
AlletherAllether
31
$endgroup$
I know this question will sound easy, but i just cant figure this out. Please help me.
How many ways are there for putting 4 pairs to the round table, so that atleast 1 pair is sitting next to each other.
combinatorics
combinatorics
asked Dec 10 '18 at 17:19
AlletherAllether
31
asked Dec 10 '18 at 17:19
AlletherAllether
31
asked Dec 10 '18 at 17:19
AlletherAllether
31
asked Dec 10 '18 at 17:19
AlletherAllether
31
asked Dec 10 '18 at 17:19
AlletherAllether
31
31
$begingroup$
How many ways can you pick a pair? How many ways can that pair be seated so that they are next to each other? Then how many ways can you seat everyone else?
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:23
$begingroup$
My 'solution' is this: i take 1 pair aside so i will always have atleast 1 pair next to each other, now for 3 pairs its 5! and now i dont know how to include that 1 pair to this.. my result is 5! + 6*2, because there are 6 places where to put that pair and *2 because they can sit like man-woman and woman-man
$endgroup$
– Allether
Dec 10 '18 at 17:27
$begingroup$
The fact that they are seated at a round table suggests to me that two seatings are equivalent if they are the same "up to rotation".
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:56
$begingroup$
One approach is to subtract the number of arrangements in which no pair is sitting next to each other from the total number of arrangements.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:55
add a comment |
$begingroup$
How many ways can you pick a pair? How many ways can that pair be seated so that they are next to each other? Then how many ways can you seat everyone else?
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:23
$begingroup$
My 'solution' is this: i take 1 pair aside so i will always have atleast 1 pair next to each other, now for 3 pairs its 5! and now i dont know how to include that 1 pair to this.. my result is 5! + 6*2, because there are 6 places where to put that pair and *2 because they can sit like man-woman and woman-man
$endgroup$
– Allether
Dec 10 '18 at 17:27
$begingroup$
The fact that they are seated at a round table suggests to me that two seatings are equivalent if they are the same "up to rotation".
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:56
$begingroup$
One approach is to subtract the number of arrangements in which no pair is sitting next to each other from the total number of arrangements.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:55
$begingroup$
How many ways can you pick a pair? How many ways can that pair be seated so that they are next to each other? Then how many ways can you seat everyone else?
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:23
$begingroup$
How many ways can you pick a pair? How many ways can that pair be seated so that they are next to each other? Then how many ways can you seat everyone else?
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:23
$begingroup$
My 'solution' is this: i take 1 pair aside so i will always have atleast 1 pair next to each other, now for 3 pairs its 5! and now i dont know how to include that 1 pair to this.. my result is 5! + 6*2, because there are 6 places where to put that pair and *2 because they can sit like man-woman and woman-man
$endgroup$
– Allether
Dec 10 '18 at 17:27
$begingroup$
My 'solution' is this: i take 1 pair aside so i will always have atleast 1 pair next to each other, now for 3 pairs its 5! and now i dont know how to include that 1 pair to this.. my result is 5! + 6*2, because there are 6 places where to put that pair and *2 because they can sit like man-woman and woman-man
$endgroup$
– Allether
Dec 10 '18 at 17:27
$begingroup$
The fact that they are seated at a round table suggests to me that two seatings are equivalent if they are the same "up to rotation".
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:56
$begingroup$
The fact that they are seated at a round table suggests to me that two seatings are equivalent if they are the same "up to rotation".
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:56
$begingroup$
One approach is to subtract the number of arrangements in which no pair is sitting next to each other from the total number of arrangements.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:55
$begingroup$
One approach is to subtract the number of arrangements in which no pair is sitting next to each other from the total number of arrangements.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A standard way is to use inclusion-exclusion principle. You count all the possible ways for a certain pair sitting next to each other, then you subtract the overcount and keep doing that until you get the correct count.
Denote the four pairs as $1,2,3,4$. For non-empty $S subseteq [4]={1,2,3,4}$,
denote $A_S$ as number of ways to put four pairs on a round table
such that pairs $iin S$ sitting next to each other.
To count $A_S$, we can see this $|S|$ pairs
as $|S|$ members in the table, plus $8-2|S|$ non-pair members, which gives
total of $8-|S|$ members on a table so there are $(7-|S|)!$ ways to
arrange that. We also need to permute members within each pair so there
are $2^{|S|}$ ways to do that. This gives
$A_S= (7-|S|)! cdot 2^{|S|}$.
Now, to use inclusion-exclusion principle, we have the final answer
as
begin{align*}
sum_{Ssubseteq A,|S| ge 1}(-1)^{|S|+1}A_S & =
sum_{Ssubseteq A, |S| ge 1}(-1)^{|S|+1}
(7-|S|)! cdot 2^{|S|}, \
& = binom{4}{1} (-1)^2 cdot 6! cdot 2 + binom{4}{2}(-1)^3 cdot 5! cdot 2^2 \
&+binom{4}{3}(-1)^4cdot 4!cdot 2^3+binom{4}{4}(-1)^5cdot 3!cdot 2^4.
end{align*}
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
A standard way is to use inclusion-exclusion principle. You count all the possible ways for a certain pair sitting next to each other, then you subtract the overcount and keep doing that until you get the correct count.
Denote the four pairs as $1,2,3,4$. For non-empty $S subseteq [4]={1,2,3,4}$,
denote $A_S$ as number of ways to put four pairs on a round table
such that pairs $iin S$ sitting next to each other.
To count $A_S$, we can see this $|S|$ pairs
as $|S|$ members in the table, plus $8-2|S|$ non-pair members, which gives
total of $8-|S|$ members on a table so there are $(7-|S|)!$ ways to
arrange that. We also need to permute members within each pair so there
are $2^{|S|}$ ways to do that. This gives
$A_S= (7-|S|)! cdot 2^{|S|}$.
Now, to use inclusion-exclusion principle, we have the final answer
as
begin{align*}
sum_{Ssubseteq A,|S| ge 1}(-1)^{|S|+1}A_S & =
sum_{Ssubseteq A, |S| ge 1}(-1)^{|S|+1}
(7-|S|)! cdot 2^{|S|}, \
& = binom{4}{1} (-1)^2 cdot 6! cdot 2 + binom{4}{2}(-1)^3 cdot 5! cdot 2^2 \
&+binom{4}{3}(-1)^4cdot 4!cdot 2^3+binom{4}{4}(-1)^5cdot 3!cdot 2^4.
end{align*}
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
$endgroup$
add a comment |
$begingroup$
A standard way is to use inclusion-exclusion principle. You count all the possible ways for a certain pair sitting next to each other, then you subtract the overcount and keep doing that until you get the correct count.
Denote the four pairs as $1,2,3,4$. For non-empty $S subseteq [4]={1,2,3,4}$,
denote $A_S$ as number of ways to put four pairs on a round table
such that pairs $iin S$ sitting next to each other.
To count $A_S$, we can see this $|S|$ pairs
as $|S|$ members in the table, plus $8-2|S|$ non-pair members, which gives
total of $8-|S|$ members on a table so there are $(7-|S|)!$ ways to
arrange that. We also need to permute members within each pair so there
are $2^{|S|}$ ways to do that. This gives
$A_S= (7-|S|)! cdot 2^{|S|}$.
Now, to use inclusion-exclusion principle, we have the final answer
as
begin{align*}
sum_{Ssubseteq A,|S| ge 1}(-1)^{|S|+1}A_S & =
sum_{Ssubseteq A, |S| ge 1}(-1)^{|S|+1}
(7-|S|)! cdot 2^{|S|}, \
& = binom{4}{1} (-1)^2 cdot 6! cdot 2 + binom{4}{2}(-1)^3 cdot 5! cdot 2^2 \
&+binom{4}{3}(-1)^4cdot 4!cdot 2^3+binom{4}{4}(-1)^5cdot 3!cdot 2^4.
end{align*}
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
$endgroup$
add a comment |
$begingroup$
A standard way is to use inclusion-exclusion principle. You count all the possible ways for a certain pair sitting next to each other, then you subtract the overcount and keep doing that until you get the correct count.
Denote the four pairs as $1,2,3,4$. For non-empty $S subseteq [4]={1,2,3,4}$,
denote $A_S$ as number of ways to put four pairs on a round table
such that pairs $iin S$ sitting next to each other.
To count $A_S$, we can see this $|S|$ pairs
as $|S|$ members in the table, plus $8-2|S|$ non-pair members, which gives
total of $8-|S|$ members on a table so there are $(7-|S|)!$ ways to
arrange that. We also need to permute members within each pair so there
are $2^{|S|}$ ways to do that. This gives
$A_S= (7-|S|)! cdot 2^{|S|}$.
Now, to use inclusion-exclusion principle, we have the final answer
as
begin{align*}
sum_{Ssubseteq A,|S| ge 1}(-1)^{|S|+1}A_S & =
sum_{Ssubseteq A, |S| ge 1}(-1)^{|S|+1}
(7-|S|)! cdot 2^{|S|}, \
& = binom{4}{1} (-1)^2 cdot 6! cdot 2 + binom{4}{2}(-1)^3 cdot 5! cdot 2^2 \
&+binom{4}{3}(-1)^4cdot 4!cdot 2^3+binom{4}{4}(-1)^5cdot 3!cdot 2^4.
end{align*}
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
$endgroup$
A standard way is to use inclusion-exclusion principle. You count all the possible ways for a certain pair sitting next to each other, then you subtract the overcount and keep doing that until you get the correct count.
Denote the four pairs as $1,2,3,4$. For non-empty $S subseteq [4]={1,2,3,4}$,
denote $A_S$ as number of ways to put four pairs on a round table
such that pairs $iin S$ sitting next to each other.
To count $A_S$, we can see this $|S|$ pairs
as $|S|$ members in the table, plus $8-2|S|$ non-pair members, which gives
total of $8-|S|$ members on a table so there are $(7-|S|)!$ ways to
arrange that. We also need to permute members within each pair so there
are $2^{|S|}$ ways to do that. This gives
$A_S= (7-|S|)! cdot 2^{|S|}$.
Now, to use inclusion-exclusion principle, we have the final answer
as
begin{align*}
sum_{Ssubseteq A,|S| ge 1}(-1)^{|S|+1}A_S & =
sum_{Ssubseteq A, |S| ge 1}(-1)^{|S|+1}
(7-|S|)! cdot 2^{|S|}, \
& = binom{4}{1} (-1)^2 cdot 6! cdot 2 + binom{4}{2}(-1)^3 cdot 5! cdot 2^2 \
&+binom{4}{3}(-1)^4cdot 4!cdot 2^3+binom{4}{4}(-1)^5cdot 3!cdot 2^4.
end{align*}
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
answered Dec 10 '18 at 23:25
TenguTengu
2,63211021
2,63211021
add a comment |
add a comment |
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$begingroup$
How many ways can you pick a pair? How many ways can that pair be seated so that they are next to each other? Then how many ways can you seat everyone else?
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:23
$begingroup$
My 'solution' is this: i take 1 pair aside so i will always have atleast 1 pair next to each other, now for 3 pairs its 5! and now i dont know how to include that 1 pair to this.. my result is 5! + 6*2, because there are 6 places where to put that pair and *2 because they can sit like man-woman and woman-man
$endgroup$
– Allether
Dec 10 '18 at 17:27
$begingroup$
The fact that they are seated at a round table suggests to me that two seatings are equivalent if they are the same "up to rotation".
$endgroup$
– Robert Thingum
Dec 10 '18 at 17:56
$begingroup$
One approach is to subtract the number of arrangements in which no pair is sitting next to each other from the total number of arrangements.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:55