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Q:- In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?

I solved it using a longer and tedious method. By the way i may also know the shorter method but i m not sure about it.

PleaseHelp
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1 Answers1

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First place the people in any order - there are $6!$ ways. Then swap each couple if necessary to put them in the right order. There are three couples, and each can be swapped or stay in the same order. Any correct ordering is thus derived from one of $8=2^3$ possible random orderings. The number of correct orderings is $\frac {6!}{2^3}$.

Mark Bennet
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  • could it seen like this :- total ways 6! now,considering couple1 half of these ways are wrong ,for couple2 half of 6!/2 ways are wrong and similarly for 3rd couple half of 6!/(22) are wrong! Hence 6!/(22*2) – PleaseHelp Sep 10 '13 at 20:42
  • @AtulGangwar That looks fine to me. It depends, of course, as does my argument, on the arrangements of each couple being independent of the others - which is something to watch out for in other similar problems. – Mark Bennet Sep 10 '13 at 20:54