If six women and six men are to be seated in a row alternately , what is the number of possible arrangements ? According to the answer itβs should be $2(6!)^2$ however I just wanted to verify what is wrong with the approach that I chose ? First seat $6$ men in a row , leaving one seat between every two men vacant. Now there are $7$ seats for the women who can be arranged in ${{7}\choose{6}}6!$ ways. Hence men and women can together be arranged in $7!6!$ ways. Why am I not getting the answer with this ? Please explain
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There are two errors in your attempt:
it doesn't necessarily seat men and women alternately, e.g. $WM-MWMWMWMWMW$
you had (originally) forgotten to permute the men, if I recall correctly.
true blue anil
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Oh yes.. I understand now.. thank you for the help! β Aditi Oct 19 '17 at 08:55
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You're welcome ! β true blue anil Oct 19 '17 at 11:30
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Your mistake is as follows:
Let $*$ denote the empty space
$$*M*M*M*M*M*M*$$
what if the women choose the following combination instead:
$$WMWMWM*MWMWMW$$
In stead, a quick correction is the women actually only has two choices, to remove the first seat (that is choosing to keep the last $6$ seats) or remove the last seat. They can then permute among themselves.
Siong Thye Goh
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