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?
 A: Let us try to answer a simpler question:
The number of ways of selecting 2 teams from 2 men and 2 women, such that each team consists of a man and a woman, is:
(a) 2! (b) 5 (c) 4 (d) 1
Discussion:
Let the two men be Ram & Shyam and the women be Radha & Sita. We can have:
(Ram, Sita); (Shyam, Radha) and (Ram, Radha); (Shyam, Sita) are the two possible combinations of a pair of teams formed satisfying the given condition.
Similarly, 15 teams can be formed using 15 men and 15 women, such that each team consists of a man and a woman, in 15! ways.
This question has appeared in JEE Main, 2015, (ONLINE CBT) and the given options were:
(a) 1120 (b) 1880 (c) 1900 (d) 1240
So, it is clear that none of the options given is 15!
The 1240 option pertains to the sum of the squares of 1 to 15, and I think it is WRONG.
A: I think it shall be as follows :
For first team :
No of ways of selection 1 man from 15 and 1 woman from 15 to form 1 st team = 15C1 X 15 C1 = 15.15= 15^2 AND
No of ways of selecting team 1 man from remaining 14 men and 1 woman from 14 women to form second team = 14C1X 14C1 = 14.14 = 14^2 AND
so on..
So you will get a series consisting of `15 ^2+14^2+13^2+.....+2^2+1^2
Which is equal to sum of squares of first 15 natural numbers =
15^3/3+15^2/2+15/6 = 1240
I think that would be the answer. 
A: The possible options given seem far, far too small and I also think 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 $15$ rooms, 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 anything could reduce the answer to something as small as $2000$. 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.
A: This question was asked in JEE Main-2015 as mentioned by S Venkateswaran,
At that time the answer was $ \displaystyle \sum_{r=1}^n r^2 $ where $n$ is the number of teams to be formed. That gives us $1240$ ,Which is wrong.
The correct answer is is actually, $15!$, there is already enough discussion about it
Another way to think is, 
First team can be selected in ${}^{15}\text{C}_1 \times {}^{15}\text{C}_1 $ 
Second team can be selected in ${}^{14}\text{C}_1 \times {}^{14}\text{C}_1 $ 
So ways of selecting 15 teams are $\dfrac{(15!)^2}{15!}=15!$
This solution is uploaded on the Test practice app by the same agency which conducts the JEE Main exam. 
A: $\color{red}{Incorrect\ Solution}$
Note: If a man or woman is a part of a team, then he or she cannot be a part of another team.
So, when we are saying:

*

*Pick 1 man out of $n$ men and 1 woman out of $n$ women, i.e., $C(n, 1) \times C(n, 1) = n^2$

*Pick 1 man out of $n - 1$ men 1 one woman out of $n - 1$ women, i.e., $C(n - 1, 1) \times C(n - 1, 1) = \left(n - 1\right)^2$

*Pick 1 man out of $n - 2$ men and 1 woman out of $n - 2$ women, i.e., $C(n - 2, 1) \times C(n - 2, 1) = \left(n - 2\right)^2$

*$...$

*Pick 1 man out of $1$ man and 1 woman out of $1$ women, i.e., $C(1, 1) \times C(1, 1) = 1^2 = 1$
This way, we get the $\color{red}{wrong}$ answer:
$$\sum_{i = 1}^{n} i^2$$
Because, we just counted how many pairs we can form out of $n$ men and $n$ women.
$\color{green}{Correct\ Solution}$
Case: Form 1 team out of 1 man and 1 woman
Suppose, M = {m1} and W = {w1}. Here, we can pick (m1, w1), so one possible way to form one team.
Case: Form 2 teams out of 2 men and 2 women
Suppose, M = {m1, m2} and W = {w1, w2}. Now we pick:

*

*(m1, w1), then we need to form teams using M = {m2} and W = {w2}, which is (m2, w2).

*(m1, w2), then we need to form teams using M = {m2} and W = {w1}, which is (m2, w1).

*(m2, w1), then we need to form teams using M = {m1} and W = {w2}, which is (m1, w2).

*(m2, w2), then we need to form teams using M = {m1} and W = {w1}, which is (m1, w1).

Therefore, the possible two possible ways to form two teams:

*

*(m1, w1) and (m2, w2)

*(m1, w2) and (m2, w1)
Notice that we have repetitive teams when solved by picking man m2, it's because when we picked the man m1 the other pair already included m2. So, we can always solve by picking one man only.
Case: Form 3 teams out of 3 men and 3 women
Suppose, M = {m1, m2, m3} and W = {w1, w2, w3}. Now we can pick:

*

*(m1, w1), then we need to form teams using M = {m2, m3} and W = {w2, w3}. It is identical problem, i.e., form two teams using two men and two women and we know that there are two teams possible:

*

*(m1, w1), (m2, w2), and (m3, w3)

*(m1, w1), (m2, w3), and (m3, w2)



*(m1, w2), two teams possible using M = {m2, m3} and W = {w1, w3}.

*(m1, w3), two teams possible using M = {m2, m3} and W = {w1, w2}.

Thus, total six possible ways to form three teams.
Generalization

*

*We first find all the possible pair of a fixed man out of $n$ men and all $n$ women. The number of ways to do this is $1 \times C(n, 1) = n$.

*Next, we find all the possible pair of a fixed man out of $n - 1$ men and remaining $n - 1$ women. The number of ways to do this is $1 \times C(n - 1, 1) = n - 1$.

*Next, we find all the possible pair of a fixed man out of $n - 2$ men and remaining $n - 2$ women. The number of ways to do this is $1 \times C(n - 2, 1) = n - 2$.

*$...$

*Next, we find all the possible pair of a fixed man out of $1$ men and remaining $1$ women. The number of ways to do this is $1 \times C(1, 1) = 1$.

Note that, we need to do all the above, so the total number of possible ways are:
$$\boxed{\left(n\right) \times \left(n - 1\right) \times \left(n - 2\right) \times \dots 1 = n!}$$
A: 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$=(15*15).(14*14)...(1*1)=(15!)^2$
