# 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)

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)

• Do you count $BG$ and $GB$ as $1$ possibility or $2$? Dec 16, 2019 at 16:18
• 2 possibilities. Dec 16, 2019 at 16:19
• Can you have $BG$ at one table, then $GB$ at another table? The first assumes you can have either all girls on the left and boys on the right or all boys on the left and girls on the right. The second assumes that each table can be different with the gender of the person on the left and right, so long as there is a boy and girl at each table. Dec 16, 2019 at 16:20
• There are many ambiguities here. It is traditional to consider each person as being distinguishable. Now, as for the tables themselves, it is unclear whether these tables are distinguishable or not. If we have an arrangement where Amy and Adam are at a table and Bob and Bernadette are at another table, this might be considered the same outcome as where Bob and Bernadette are at a table and Amy and Adam are at another table or they might have been considered different. If, say, at one table they are serving fish and at another table they are serving chicken, if Amy has an allergy it matters Dec 16, 2019 at 16:32
• Then, it also matters whether we consider arrangement within a table to matter. If we have an arrangement where Amy and Adam are at a table, and so on, is this necessarily a different arrangement than an arrangement where Adam and Amy are at a table... Maybe one of the chairs at the table is 20 feet high and the other chair is very short. If Amy is scared of heights, it matters to her whether she is in the 20 foot chair or not. Dec 16, 2019 at 16:34

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:

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

• Perfect, thank you! Dec 16, 2019 at 16:47

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.

• OK. So, can I think about it as dividing into pairs(6!6!) and then choosing a table? Dec 16, 2019 at 16:33
• Yes. That would be a good strategy here. Dec 16, 2019 at 16:35
• @Combinatoric It is not clear what you mean by dividing into pairs (6!6!) and then choosing a table... 6! is choosing which boy (or girl) sits at which table. Dec 16, 2019 at 16:39
• @InterstellarProbe can you find another way to explain it? I want to make sure I really get it. Dec 16, 2019 at 16:40
• @Combinatoric assuming tables are distinguishable, for table A, pick a boy ($6$ options) then pick a girl ($6$ options). Next, for table B pick a boy from those remaining ($5$ options) and then pick a girl from those remaining ($5$ options). Continue until all tables have the people picked, giving $6\cdot 6\cdot 5\cdot 5\cdots 1\cdot 1 = 6!\cdot 6!$ options for distributing the people. If we then care about which chair was used for each person at the tables themselves, for table A pick if the boy sat in the high chair or the low chair and let the girl sit in the other, for $2$ options, etc. Dec 16, 2019 at 16:42