Question 342369
<pre><b>
First we find the numerator of the probability.

The number of ways we can have this arrangement

M W M W M W M W M W

For each of the 5! ways to place the men there are 5! ways to place the
women. That's 5!*5! or (5!)²

But we could also have this arrangement:

W M W M W M W M W M

which is another (5!)²

So the numerator, which is the number of ways to have men and women in
alternate seats, is 2(5!)²

The denominator is the number of ways anybody can sit anywhere, which
is 10!

So the probability is

{{{( 2(5!)^2)/10!=(2(5*4*3*2*1)(5*4*3*2*1))/(10*9*8*7*6*5*4*3*2*1)=


(2(cross(5)*cross(4)*cross(3)*cross(2)*1)(cross(5)*cross(4)*cross(3)*cross(2)*1))/(cross(10)*cross(9)*cross(8)*7*cross(6)*cross(5)*cross(4)*cross(3)*cross(2)*1) =  2/(2*3*2*7*3)=cross(2)/(cross(2)*3*2*7*3)=1/126}}}
                        {{{2}}} {{{3}}}{{{2}}}  {{{3}}}


Edwin</pre>