# Arrangement of people in a row

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 it’s 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

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.

• Okay ! I understand now . Thank you for the help. – Aditi Oct 19 '17 at 6:40

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.

• Oh yes.. I understand now.. thank you for the help! – Aditi Oct 19 '17 at 8:55
• You're welcome ! – true blue anil Oct 19 '17 at 11:30