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Mathematics 10 Online
OpenStudy (dls):

There are two rows,one behind other,of 5 chairs each.Five couples are to be seated, number of arrangements such that no husband sits in front or behind his wife

OpenStudy (anonymous):

impossibruuu level:asian

OpenStudy (dls):

:(

OpenStudy (kropot72):

The permutations theorem dealing with classes of different things could apply here. If n given things can be divided into c classes of alike things differing from class to class, then the number of permutations of these things taken all at a time is: \[\frac{n!}{n _{1}!n _{2}!....n _{c}!}where(n _{1}+n _{2}+.....+n _{c}=n)\] The classes in the question are the 5 different couples. So the number of seating arrangements is: \[\frac{10!}{2!2!2!2!2!}\]

OpenStudy (dls):

ans=wrong

OpenStudy (dls):

it gives 113400 butits190080

OpenStudy (anonymous):

told u imposssibru lvl: stephen hawking

OpenStudy (dls):

lol

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