Question

Find the number of ways in which $6$ persons out of $5$ men & $5$ women can be seated at a round table such that $2$ men are never together.

Answer 1

Lets first place the men $M$.

Here $\ast$ indicates the linker of round table

$\ast M-M-M-M-M\ast$

which is in $(5-1)!$ ways

So we have to place the women in between the men which is on the $5$ empty seats (4 -'s and 1 linker i.e * )

So, $5$ women can sit on $5$ seats in $\left(5\right)!$ ways or

1st seat in $5$ ways

2nd seat $4$ ways

3rd seat $3$ ways

4th seat $2$ ways

5th seat $1$ ways

i.e $5\ast 4\ast 3\ast 2\ast 1$ ways

So, the answer is $5!\times 4!=2880$