# In how many ways 3 couples can be arranged in a line such that each husband is ahead of his wife?

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.

• You are only considering the case where the husband is directly in front of the wife. The husband may be 2 people behind or even further and the condition in the question is still satisfied – Harry Alli Sep 22 '17 at 13:11
• @HarryAlli I assume you mean "2 people ahead" but yes, I think your comment is correct – Gregory J. Puleo Sep 22 '17 at 13:12
• @HarryAlli Thanks, I interpreted it wrongly. – AMAN Sep 22 '17 at 17:26

$${6\choose2}{4\choose2}{2\choose2}=15\times6\times1=\mathbf{90}$$