21The 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
21
Shubhodip
·2012-04-16 23:23:31
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.
9choose 1 couple either 2 men or 2 women=
6C15C25C0
now choose 1 couple one man and another woman keeping in mind that they are not married
so
6C15c14C1
we choose men 5c1 and then choose from the four women other than his wife 4C1
total=6C15C25C0+6C15C25C0+6C15c14C1=240