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Differing computations for combinatorics question
$begingroup$
How many seating arrangements are there with 6 boys and 6 girls if the table is hexagonal with 2 seats on each side, out of which one must be occupied by a boy and the other by a girl?
I approached this from 2 different perspectives:
First approach: We must have a boy-girl pair for each side of the hexagon. The total number of possible pairs is $6 times 6 = 36$. Out of these $36$ possible pairs, we need to choose 6 pairs to arrange around the table, after which we have to permute the pairs (treating each pair as a single entity). We get $C(36,6) times 6!$. Then, considering the number of permutations within each pair and that the table has 6 sides, the final answer is $frac{C(36,6) times 6! times 2!}{6}$.
Second approach: We first consider the possibilities in a straight line. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls. We then have to consider the permutations within each pair and that we are arranging them around a table with 6 sides, so the final answer is $frac{6! times 6! times 2^6}{6}$.
The line of thought I applied in both approaches felt completely reasonable, but they give differing answers, why is that so? Have I overcounted/undercounted somewhere?
combinatorics discrete-mathematics permutations
asked Jan 31 at 2:10
uznamuznam
437
$endgroup$
add a comment |
$begingroup$
How many seating arrangements are there with 6 boys and 6 girls if the table is hexagonal with 2 seats on each side, out of which one must be occupied by a boy and the other by a girl?
I approached this from 2 different perspectives:
First approach: We must have a boy-girl pair for each side of the hexagon. The total number of possible pairs is $6 times 6 = 36$. Out of these $36$ possible pairs, we need to choose 6 pairs to arrange around the table, after which we have to permute the pairs (treating each pair as a single entity). We get $C(36,6) times 6!$. Then, considering the number of permutations within each pair and that the table has 6 sides, the final answer is $frac{C(36,6) times 6! times 2!}{6}$.
Second approach: We first consider the possibilities in a straight line. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls. We then have to consider the permutations within each pair and that we are arranging them around a table with 6 sides, so the final answer is $frac{6! times 6! times 2^6}{6}$.
The line of thought I applied in both approaches felt completely reasonable, but they give differing answers, why is that so? Have I overcounted/undercounted somewhere?
combinatorics discrete-mathematics permutations
asked Jan 31 at 2:10
uznamuznam
437
$endgroup$
add a comment |
$begingroup$
How many seating arrangements are there with 6 boys and 6 girls if the table is hexagonal with 2 seats on each side, out of which one must be occupied by a boy and the other by a girl?
I approached this from 2 different perspectives:
First approach: We must have a boy-girl pair for each side of the hexagon. The total number of possible pairs is $6 times 6 = 36$. Out of these $36$ possible pairs, we need to choose 6 pairs to arrange around the table, after which we have to permute the pairs (treating each pair as a single entity). We get $C(36,6) times 6!$. Then, considering the number of permutations within each pair and that the table has 6 sides, the final answer is $frac{C(36,6) times 6! times 2!}{6}$.
Second approach: We first consider the possibilities in a straight line. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls. We then have to consider the permutations within each pair and that we are arranging them around a table with 6 sides, so the final answer is $frac{6! times 6! times 2^6}{6}$.
The line of thought I applied in both approaches felt completely reasonable, but they give differing answers, why is that so? Have I overcounted/undercounted somewhere?
combinatorics discrete-mathematics permutations
asked Jan 31 at 2:10
uznamuznam
437
$endgroup$
How many seating arrangements are there with 6 boys and 6 girls if the table is hexagonal with 2 seats on each side, out of which one must be occupied by a boy and the other by a girl?
I approached this from 2 different perspectives:
First approach: We must have a boy-girl pair for each side of the hexagon. The total number of possible pairs is $6 times 6 = 36$. Out of these $36$ possible pairs, we need to choose 6 pairs to arrange around the table, after which we have to permute the pairs (treating each pair as a single entity). We get $C(36,6) times 6!$. Then, considering the number of permutations within each pair and that the table has 6 sides, the final answer is $frac{C(36,6) times 6! times 2!}{6}$.
Second approach: We first consider the possibilities in a straight line. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls. We then have to consider the permutations within each pair and that we are arranging them around a table with 6 sides, so the final answer is $frac{6! times 6! times 2^6}{6}$.
The line of thought I applied in both approaches felt completely reasonable, but they give differing answers, why is that so? Have I overcounted/undercounted somewhere?
combinatorics discrete-mathematics permutations
combinatorics discrete-mathematics permutations
asked Jan 31 at 2:10
uznamuznam
437
asked Jan 31 at 2:10
uznamuznam
437
asked Jan 31 at 2:10
uznamuznam
437
asked Jan 31 at 2:10
uznamuznam
437
asked Jan 31 at 2:10
uznamuznam
437
437
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let your boys be capital letters $A,B,C,dots$ and the girls be lowercase $a,b,c,dots$. In your first approach, the $6times 6=36$ possible pairs include ${(A,a),(A,b),(A,c)dots,(B,a),(B,b)dots}$ When choosing six of these pairings without further stipulation as you had, you might have chosen pairings where one or more of the people are repeated... for example the six pairings which all used boy $A$.
The second approach is the correct one, however I would suggest avoiding "division by symmetry" arguments where possible as it makes several people uncomfortable. That isn't to say that it is wrong, just that if it can be avoided it usually leads to a "nicer" proof.
Here, we can let boy $A$ take an edge around the table and then pick whether he sits in the left seat at the edge or the right seat at the edge. For this, there are only $2$ options that really mattered so far, the left or the right. Which edge was chosen does not at all matter.
From here, we can now have the rest of the boys pick which edge in relation to the first boy they sit at, and same too with the girls, then picking whether the boy sits at the left or at the right on the edge, for a total number of arrangements being
$$5!times 6!times 2^6$$
the same answer as you got in the second approach.
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Each side has exactly 2 possibilities. Either the boy is clockwise from the girl or he is counterclockwise from the girl. Each side is independent of all of the other sides, so there are a total of $2^6=64$ possible arrangements if the boys and girls are indistinguishable from one another.
Of course, the boys are girls are distinguishable. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls (again, the arrangements are independent), so the total number of possible arrangements is $6! times 6! times 2^6$.
Your second solution assumes (by dividing by 6) that solutions are equivalent if they differ by a rotation. Your first solution is wrong because after choosing the first boy-girl pair, you no longer have 35 remaining possibilities.
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let your boys be capital letters $A,B,C,dots$ and the girls be lowercase $a,b,c,dots$. In your first approach, the $6times 6=36$ possible pairs include ${(A,a),(A,b),(A,c)dots,(B,a),(B,b)dots}$ When choosing six of these pairings without further stipulation as you had, you might have chosen pairings where one or more of the people are repeated... for example the six pairings which all used boy $A$.
The second approach is the correct one, however I would suggest avoiding "division by symmetry" arguments where possible as it makes several people uncomfortable. That isn't to say that it is wrong, just that if it can be avoided it usually leads to a "nicer" proof.
Here, we can let boy $A$ take an edge around the table and then pick whether he sits in the left seat at the edge or the right seat at the edge. For this, there are only $2$ options that really mattered so far, the left or the right. Which edge was chosen does not at all matter.
From here, we can now have the rest of the boys pick which edge in relation to the first boy they sit at, and same too with the girls, then picking whether the boy sits at the left or at the right on the edge, for a total number of arrangements being
$$5!times 6!times 2^6$$
the same answer as you got in the second approach.
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Let your boys be capital letters $A,B,C,dots$ and the girls be lowercase $a,b,c,dots$. In your first approach, the $6times 6=36$ possible pairs include ${(A,a),(A,b),(A,c)dots,(B,a),(B,b)dots}$ When choosing six of these pairings without further stipulation as you had, you might have chosen pairings where one or more of the people are repeated... for example the six pairings which all used boy $A$.
The second approach is the correct one, however I would suggest avoiding "division by symmetry" arguments where possible as it makes several people uncomfortable. That isn't to say that it is wrong, just that if it can be avoided it usually leads to a "nicer" proof.
Here, we can let boy $A$ take an edge around the table and then pick whether he sits in the left seat at the edge or the right seat at the edge. For this, there are only $2$ options that really mattered so far, the left or the right. Which edge was chosen does not at all matter.
From here, we can now have the rest of the boys pick which edge in relation to the first boy they sit at, and same too with the girls, then picking whether the boy sits at the left or at the right on the edge, for a total number of arrangements being
$$5!times 6!times 2^6$$
the same answer as you got in the second approach.
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Let your boys be capital letters $A,B,C,dots$ and the girls be lowercase $a,b,c,dots$. In your first approach, the $6times 6=36$ possible pairs include ${(A,a),(A,b),(A,c)dots,(B,a),(B,b)dots}$ When choosing six of these pairings without further stipulation as you had, you might have chosen pairings where one or more of the people are repeated... for example the six pairings which all used boy $A$.
The second approach is the correct one, however I would suggest avoiding "division by symmetry" arguments where possible as it makes several people uncomfortable. That isn't to say that it is wrong, just that if it can be avoided it usually leads to a "nicer" proof.
Here, we can let boy $A$ take an edge around the table and then pick whether he sits in the left seat at the edge or the right seat at the edge. For this, there are only $2$ options that really mattered so far, the left or the right. Which edge was chosen does not at all matter.
From here, we can now have the rest of the boys pick which edge in relation to the first boy they sit at, and same too with the girls, then picking whether the boy sits at the left or at the right on the edge, for a total number of arrangements being
$$5!times 6!times 2^6$$
the same answer as you got in the second approach.
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
$endgroup$
Let your boys be capital letters $A,B,C,dots$ and the girls be lowercase $a,b,c,dots$. In your first approach, the $6times 6=36$ possible pairs include ${(A,a),(A,b),(A,c)dots,(B,a),(B,b)dots}$ When choosing six of these pairings without further stipulation as you had, you might have chosen pairings where one or more of the people are repeated... for example the six pairings which all used boy $A$.
The second approach is the correct one, however I would suggest avoiding "division by symmetry" arguments where possible as it makes several people uncomfortable. That isn't to say that it is wrong, just that if it can be avoided it usually leads to a "nicer" proof.
Here, we can let boy $A$ take an edge around the table and then pick whether he sits in the left seat at the edge or the right seat at the edge. For this, there are only $2$ options that really mattered so far, the left or the right. Which edge was chosen does not at all matter.
From here, we can now have the rest of the boys pick which edge in relation to the first boy they sit at, and same too with the girls, then picking whether the boy sits at the left or at the right on the edge, for a total number of arrangements being
$$5!times 6!times 2^6$$
the same answer as you got in the second approach.
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
answered Jan 31 at 2:36
JMoravitzJMoravitz
48.7k43988
48.7k43988
add a comment |
add a comment |
$begingroup$
Each side has exactly 2 possibilities. Either the boy is clockwise from the girl or he is counterclockwise from the girl. Each side is independent of all of the other sides, so there are a total of $2^6=64$ possible arrangements if the boys and girls are indistinguishable from one another.
Of course, the boys are girls are distinguishable. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls (again, the arrangements are independent), so the total number of possible arrangements is $6! times 6! times 2^6$.
Your second solution assumes (by dividing by 6) that solutions are equivalent if they differ by a rotation. Your first solution is wrong because after choosing the first boy-girl pair, you no longer have 35 remaining possibilities.
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
$endgroup$
add a comment |
$begingroup$
Each side has exactly 2 possibilities. Either the boy is clockwise from the girl or he is counterclockwise from the girl. Each side is independent of all of the other sides, so there are a total of $2^6=64$ possible arrangements if the boys and girls are indistinguishable from one another.
Of course, the boys are girls are distinguishable. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls (again, the arrangements are independent), so the total number of possible arrangements is $6! times 6! times 2^6$.
Your second solution assumes (by dividing by 6) that solutions are equivalent if they differ by a rotation. Your first solution is wrong because after choosing the first boy-girl pair, you no longer have 35 remaining possibilities.
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
$endgroup$
add a comment |
$begingroup$
Each side has exactly 2 possibilities. Either the boy is clockwise from the girl or he is counterclockwise from the girl. Each side is independent of all of the other sides, so there are a total of $2^6=64$ possible arrangements if the boys and girls are indistinguishable from one another.
Of course, the boys are girls are distinguishable. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls (again, the arrangements are independent), so the total number of possible arrangements is $6! times 6! times 2^6$.
Your second solution assumes (by dividing by 6) that solutions are equivalent if they differ by a rotation. Your first solution is wrong because after choosing the first boy-girl pair, you no longer have 35 remaining possibilities.
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
$endgroup$
Each side has exactly 2 possibilities. Either the boy is clockwise from the girl or he is counterclockwise from the girl. Each side is independent of all of the other sides, so there are a total of $2^6=64$ possible arrangements if the boys and girls are indistinguishable from one another.
Of course, the boys are girls are distinguishable. There are $6!$ ways to arrange the boys and $6!$ ways to arrange the girls (again, the arrangements are independent), so the total number of possible arrangements is $6! times 6! times 2^6$.
Your second solution assumes (by dividing by 6) that solutions are equivalent if they differ by a rotation. Your first solution is wrong because after choosing the first boy-girl pair, you no longer have 35 remaining possibilities.
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
answered Jan 31 at 2:35
Robert ShoreRobert Shore
3,611324
3,611324
add a comment |
add a comment |
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