In how many ways can we split 6 boys and 6 girls to 6 tables such that in every table there's a boy and girl (only 2 seats per table)

Question

In how many ways can we split 6 boys and 6 girls to 6 tables such that in every table there's a boy and girl (only 2 seats per table)?

I saw this post and I think it's kind of the same of my question

In how many ways if no two people of the same sex are allowed to sit together?

And the answer is $2*(6!)^2$

But then I saw this one (e) -

Discrete Mechatronics - sequences with repeats and no repeats

Which says the answer is $2^6*(6!)^2$

I would like to know which one is correct in my case and why? (like what are the differences)

Answer 1

Fix a table. Choose a boy and a girl for the table. Choose an order for the boy and girl.

You have six boys and six girls to choose from. Then, you can order them in two ways. So, this is $6\times 6\times 2$ ways to arrange one boy and one girl at the first table.

For the second table, you have five boys left, five girls left, and two ways to arrange then, entirely independent from the arrangement of the first table (the only dependency is that you have one fewer boy and one fewer girl to choose from).

So, there are $5\times 5\times 2$ ways to arrange one boy and one girl at the second table (after a boy and girl were already arranged at the first table).

For the third table, you have $4\times 4\times 2$ ways to arrange them. For the fourth table, you have $3\times 3 \times 2$ ways to arrange them. For the fifth table, you have $2\times 2\times 2$ ways to arrange them. For the last table, you have $1\times 1\times 2$ ways to arrange them.

Because each table is arranged independently from the arrangements of the previous tables, you can apply the product principle. So, the total is:

$$(6\cdot 6\cdot 2)(5\cdot 5\cdot 2)(4\cdot 4\cdot 2)(3\cdot 3\cdot 2)(2\cdot 2\cdot 2)(1\cdot 1\cdot 2) = 2^6(6!)^2$$

Answer 2

In the first reference, the boys and girls are sitting in a line with no two people of the same gender are allowed to sit together. So there can be two cases. First in which a boy sits first and the second when a girl sits first. While in your case as you've mentioned in the comments there can be two possibilities for each table GB or BG. Thus you have a multiplier of 2 for each table along with every permutation of boys and girls.

In other words, $2*(6!)^2$ is equivalent to the case when each table should have the same layout i.e either all the table should be BG or GB, while $2^6*(6!)^2$ is the case where each table can BG or GB independent of the other.