In how many ways 3 couples can be arranged in a line such that each husband is ahead of his wife?
a) $81$
b) $90$
c) $110$
d) $125$
e) $132$
I tried to solve this, as 3 couples are there and they should be together, number of permutations are $3!$ and as it is fixed that each husband will be ahead of his wife, no further permutation is required for husband and wife.
Thus, my answer is $ 3! = 6$, but it is nowhere close to any of the option.
Best Answer
We firstly pick two positions for the first couple to occupy, putting the man in front of the woman, then we pick another two for the second couple to occupy, again putting the man in front. This will leave two positions for the third couple to occupy.
$${6\choose2}{4\choose2}{2\choose2}=15\times6\times1=\mathbf{90}$$
Thus our answer is b)