Find the number of ways to form 15 teams out of 15 men and 15 women. In how many ways can 15 teams be formed, each consisting of a man and a woman, from 15 men and 15 women.

This looks like the same problem as finding the number of bijective functions from a set $A$ to $B$, both containing 15 elements each.

So the answer should be $15!$ which is not correct.

Edit: This is not a legitimate question though, could somebody try thinking about what constraints need to be added, so that the answer is close to the options. Yes, I mean 'correct the question'; it uses very rough language and is probably not what the options imply.

Edit: Let's try breaking the problem. What would be the number of ways to form 3 teams, each consisting of a man and a woman?
The possible options given seem far, far too small. I too believe the answer is $15!$.

There are many ways of thinking about it, but perhaps the easiest is to remove the identities of the men and line them up as a series of $15$ bins (hm, maybe 'rooms' instead here), in to which we place the women. Since we assume we only care which woman a man is with, we only care which room a woman is assigned to, the order and labelling of the rooms doesn't matter. There are then, as usual, $15$ options then $14$ options and so on, giving $15!$.

I am struggling to see how any interpretation of the problem could reduce the answer to something as small as roughly $2000$. It seems to me this would require further labels to be removed, for example treating a load of the men as being called Dave, and saying they're the same (no disrespect to Daves).
There are many ways to approach this problem. One is that- the first man will have 15 choices. Then the 2$^{nd}$ one will have 14 and like this till 1. So total choices for men-$15!$ 
Now in the same way total choices for women will be $15!$. 
So total ways= $15!.15!=(15!)^2$

One more way to approach this problem is- For the first team there will be $(15*15)$ choices. For  the second one- $(14*14)$ , till $(1*1)$ 
So total ways$=(15*15).(14*14)...(1*1)=(15!)^2$