Dance couples permutation question Need some help with this question...

At a party $6$ boys and $6$ girls dance together. Assuming that the classical dance is performed, in which one couple (one boy and one girl), how many couples can perform together?

I know the answer is $6!$ because you take just one group (boys or girls) to assess but I don't understand the logic behind it. If someone could enlighten me I'll be very grateful.
 A: To give an intuitive angle to the problem, imagine that a single boy can dance with any of the 6 girls. When this happens, the next boy can dance with any of the 5 remaining girls, the next boy any of the 4 remaining girls, and so on, so we have the following ways of arranging the dance partners:
$$6\times 5 \times 4 \times 3 \times 2 \times 1 = 6!$$
This is how we get $6!$ as our answer. I hope this helps you to understand, if you still have questions, feel free to comment and I will do my best to answer them.
A: The question is phrased poorly; taken literally, with 6 boys and 6 girls, the answer to "How many couples can perform together?" is "Six." You can't have more than that many couples, because you would need more people, and you can certainly have six couples performing at the same time.
However, it seems clear that the question is meant to be something along the lines of:

In how many different ways can we pair the boys and the girls to make six couples that will all dance at the same time?

Now, since the order in which we list the pairs doesn't matter, we just care about what girl goes with what boy. So we can arrange the girls on a line and keep them fixed, and just see in how many ways we can "shuffle" the boys to pair them up with the girls: indeed, each of these will lead to a different set of six pairings. And given any set of six pairings, we can re-list them so that the girls appear in the order we selected. So we just need to count them under this assumption (the girls are ordered, say alphabetically by name, and we just need to order the boys to decide who dances with which girl).  Viewed in this light, the question is equivalent to asking in how many different ways we can order the six boys.
There are six possibilities for the boy who goes first. That leaves five for the boy who goes second, four for the boy who goes third, etc. We are computing the permutations of six objects, and the answer is simply $6\times 5\times 4\times 3\times 2\times 1 = 6!$. 
A: What does the term "How many couples can perform together?" mean.  If it means how many distinct ways can the boys and girl be matched up then it is 6!  But if it just means how many pairing of boys and girls can there be then the answer is 6x6=36 because each boy can be matched up with any one of the 6 girls.  I think the question is ambiguous.
