Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?
I have no OA..
Please explain.
Men & Women around circular table
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imo 6!*7!
7 men can be seated in 6! (circular permutation (n-1)!)
in b/n them 7 women can be seated in 7! ways.....
so 6!*7!
7 men can be seated in 6! (circular permutation (n-1)!)
in b/n them 7 women can be seated in 7! ways.....
so 6!*7!
It does not matter how many times you get knocked down , but how many times you get up
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ok, how do we make sure that in such an arrangement, no two women sit next to each other?
my thinking was (though I could never take it to the end) to subtract number of arrangements where two women sit together from the total number of arrangements. the way we normally do it.
total number of ways that 14 people can be seated around a table is 13! (let us fix one woman and rotate other people around her)
now consider that of other six women two sit next to each other. number of arrangements in this case is 2*12!+2*12!+2*12! = 6*12! (there can be three pairs of women)
next what we have to take into account that two women can sit adjacent to a woman that we fixed from either left or right. thus we will have 1 more pair from the left and from the right. and here is where i stumble. any thoughts?
thank you!
my thinking was (though I could never take it to the end) to subtract number of arrangements where two women sit together from the total number of arrangements. the way we normally do it.
total number of ways that 14 people can be seated around a table is 13! (let us fix one woman and rotate other people around her)
now consider that of other six women two sit next to each other. number of arrangements in this case is 2*12!+2*12!+2*12! = 6*12! (there can be three pairs of women)
next what we have to take into account that two women can sit adjacent to a woman that we fixed from either left or right. thus we will have 1 more pair from the left and from the right. and here is where i stumble. any thoughts?
thank you!
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in these type of Q we need to place first the group without any condition ....that is the men here.caspermonday wrote:ok, how do we make sure that in such an arrangement, no two women sit next to each other?
my thinking was (though I could never take it to the end) to subtract number of arrangements where two women sit together from the total number of arrangements. the way we normally do it.
total number of ways that 14 people can be seated around a table is 13! (let us fix one woman and rotate other people around her)
now consider that of other six women two sit next to each other. number of arrangements in this case is 2*12!+2*12!+2*12! = 6*12! (there can be three pairs of women)
next what we have to take into account that two women can sit adjacent to a woman that we fixed from either left or right. thus we will have 1 more pair from the left and from the right. and here is where i stumble. any thoughts?
thank you!
which is 6!
after placing them there are exactly seven places in b/n them where the women can be seated satisfying the given condition that no two women are placed next to each other(as there is a men in b/n them).
this can be done in 7!
tot 6!*7!
It does not matter how many times you get knocked down , but how many times you get up
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