SOLUTION: can you answer this and I will be greatful Ten women and their husbands are to be seated on two sides of a long rectangular table, ten on each side. How many possible seating arra

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Question 388463: can you answer this and I will be greatful
Ten women and their husbands are to be seated on two sides of a long rectangular table, ten on each side. How many possible seating arrangements are there if
a) all the women sit on the same side?
b) women and men alternate along the able and every women sits opposite a man?
c) everyone sits opposite of their spouse?

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
can you answer this and I will be greatful
Ten women and their husbands are to be seated on two sides of a long rectangular table, ten on each side. How many possible seating arrangements are there if
a) all the women sit on the same side?
There are 2 possible configurations they can be placed, gender-wise

WWWWWWWWWW     MMMMMMMMMM
MMMMMMMMMM  or WWWWWWWWWW

We have 10 women to POSITION 10 ways, that 10 Position 10 or 10P10 or 10! 
For each of those 10! ways to position the women on the left, there are
10! ways to position the 10 men on the right.  That's (10!)².

Then for each of the 10! ways to place the women, there are 10! ways
to place the men.  So for each of the above seating types there are (10!)²
ways.

Since there are two kinds of configurations, the total number of ways is 
twice that, or 2(10!)² or about 26336378880000 ways.


b) women and men alternate along the table and every women sits opposite a man?

There are 2 possible ways they can be placed, gender-wise

WMWMWMWMWM     MWMWMWMWMW
MWMWMWMWMW  or WMWMWMWMWM

Each one the 10 women can be positioned 10 ways, that's 10P10 or 10!.
For each of those 10! ways to place the women, there are also 10P10 ways
to place the men.  So for each of the above two seating types there are 
(10!)² ways.

Since there are two kinds of seating types gender-wise, the total number of
ways is twice that, or 2(10!)² or about 26336378880000 ways.


c) everyone sits opposite of their spouse?

We can choose the leftmost couple any of 10 ways and there are 2 ways to
place this couple, so that's 10*2 or 20 ways to choose and seat a left-most
couple.

For each of the 10*2 ways to place the 1st (leftmost) couple, we can choose the
2nd couple just to the right of the 1st couple any of 9 ways and there are 2
ways to place this couple, so that's 9*2 or 18 ways to place the second couple
from the left.  So that's a total of 10*2*9*2 ways to place the first two
couples on the far left.
 
We can continue this process, and we will end up with:

10*2*9*2*8*2*7*2*6*2*5*2*4*2*3*2*2*2*1*2 =  10!*210 = 3715891200 ways.

Edwin