Permutations

The following works under the assumption that there are 6 men and 6 women and each men is married to exactly one woman, and each woman is married to exactly one man. (*)

First we need to select a couple, which can be done in 6 ways.

After that we may select two men, from the remaining (There wont be anymore married couple by (*) ) in C(5,2) ways.

We can also select two women, from the remaining (There wont be anymore married couple by (*) ) in C(5,2) ways.

We can select two men and one women too. If we select the men first it can be done in 5 ways and selecting the woman can only be done in 4 ways. (since we cann't select his unique wife ... )

So 20 ways.

Now after selecting a couple the rest can be taken in C(5,2) + C(5,2) + 20 =40 ways.

So TOTAL WAY : = 6* 40 = 240

Oh btw you asked for fallacy in ur solution

Its just that you are counting one non married pair twice.

First u are selecting a married couple (in 6 ways) and then selecting one person from the rest, and then selecting someone which is not his/her wife/husband

But in this way one pair appears two times.

Let the pair be (a,b). If you take a first and b later then it appear once.

and again you are considering a case when you take b first and a later. But here we are counting unordered pairs, so these two cases re same. So we need to divide it by 2.