gmat club prob

Total no. of selection = 12C4
To find 4 people such that none is a married couple, select 4 couples and a single person from each.
Thus 6C4 (=selection of 4 couples) * 2^4
Thus prob = 6C4 *16/12C4 = 15*16*24/12*11*10*9 = 16/33
IMO C

You can also use the slot method:

1st pick: can be 12 out of 12 people, so probability is 1
2nd pick: there are 11 people left, but 1 of them is the partner of the 1st pick, so probability is 10/11
3nd pick: there are 10 people left, but 2 of them are partners of the first 2 picks, so probability is 8/10
4th pick: there are 9 people left, but 3 of them are partners of the first 3 picks, so probability is 6/9.

1*(10/11)*(8/10)*(6/9) = 16/33

cans wrote: To find 4 people such that none is a married couple, select 4 couples and a single person from each.
Thus 6C4 (=selection of 4 couples) * 2^4

did'nt get this part cans. can u kindly elaborate ?

To find 4 people such that none is a married couple, select 4 couples and a single person from each.
Thus 6C4 (=selection of 4 couples) * 2^4

We have to select 4 people such that no 2 form a couple. My method is select 4 couples first = 6C4 ways
Now from each couple we can select only one person (if we select both, then condition will not be met)
Thus from 1st couple 2 ways (either girl or guy)
from 2nd couple 2 ways again
from 3rd - 2 ways
from 4th - 2 ways
Thus total of 6C4*2*2*2*2 = 6C4*2^4.
I hope you get it now.

yes, i do,,this question was a bit tough I guess.


ok, just to make sure :
Thus from 1st couple 2 ways (either girl or guy)
from 2nd couple 2 ways again
from 3rd - 2 ways
from 4th - 2 ways

what about the 5th and 6th couple ?

what about the 5th and 6th couple ?

We are not concerned about 5th and 6th couple as we have to select only 4 people, and initially we selected 4 couples from 6 (remember 6C4 done initially)