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?

• I'm lost. I also think $15!$ should be the answer. – MonkeyKing Apr 12 '15 at 20:13
• Well it depends on whether or not each man and woman can only be part of a single team or if a man (resp. woman) can be part of more than one team. – DAS Apr 12 '15 at 20:16
• Also, since you say $15!$ is wrong, probably you have the correct answer? What about showing us so that we can try to find out what the question means. Sometimes such question is just about understanding what the question precisely means. – MonkeyKing Apr 12 '15 at 20:19
• Can two "different" teams consist of the exact same couple? I.e., are we to think of each team as having a distinct "name," such as some task it is to perform? – Barry Cipra Apr 12 '15 at 20:54
• Unless there are more constraints, the answer is $15!$. – Jean-Claude Arbaut Apr 15 '15 at 7:55

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).

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.

• Generally, if you want to count the number of ways of making a series of choices, you multiply the number of ways to make each choice, not add. – Alex Zorn May 1 '15 at 18:28
• You wrote "AND" but the processes do not occur independently but rather occur sequentially. Since we have to form 15 teams, we form one team , "THEN" the second team, "THEN" the third team and so on. So we have to multiply. But that was the correct answer. So this was what was required and the question does not clarify it. – Shubham May 14 '15 at 15:47

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.

• Thanks for the answer. I'd encourage you to expand by answering newer questions. – Tianlalu Nov 22 '18 at 11:05

There are $15!$ ways to distribute the men onto the teams. There are also $15!$ ways to distribute the women. Therefore, the answer would be $$15!\cdot15! = (15!)^2$$

• This is the way of distributing them onto 15 distinct teams (i.e. team A, team B, team C, ...). It is often implied that we cannot tell the difference between team names, and all we care about is who is partnered with who. – JMoravitz Apr 12 '15 at 20:07
• We need not consider both arrangements. Once the men( or women) are distributed, we should get all distinct teams. I don't remember the answer but it was very much less than $15!$. So $15! \times 15!$ must be orders of magnitude larger! – Shubham Apr 12 '15 at 20:10

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$=(15*15).(14*14)...(1*1)=(15!)^2$