Question 1207318
<font color=black size=3>
Answers:
(a) <font color=red>1/6</font>
(b) <font color=red>1/396</font>


----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------


Explanation for Part (a)


The first 12 letters of the English alphabet are A through L.


Let A and B represent the two people that make up this particular couple that want to sit together.


Let X represent their place in any given permutation. 
The sequence C,D,X,E,F,G,H,I,J,K,L is one such example.
We'll replace X with either A,B or B,A.


Doing it this way guarantees that A & B stick together.
i.e. it guarantees that A & B are next door neighbors.


There are initially 12 letters.
Removing A & B temporarily drops things to 12-2 = 10 letters.
Temporarily introducing X bumps it up to 10+1 = 11 letters.


There are 2*11! ways to arrange the 12 people so that A,B sit together in either order. 
The exclamation mark indicates factorial.
The 11! factorial term is the number of ways to arrange the 11 letters involving X (but not A,B just yet), and the 2 out front is to consider the two cases when A,B happens or B,A happens.


This is out of 12! ways to arrange all 12 people regardless if that particular couple is sitting together or not.


So we have:
2*11! = number of ways to arrange people so A,B stick together
12! = number of permutations total


Divide the two items mentioned:
(2*11!)/(12!)
(2*11!)/(12*11!)
2/12
<font color=red>1/6</font> is the answer to part (a).
Notice the 11! terms cancel which is really convenient.


Another thing to note is if you had n people then 2/n is the probability that one particular couple sticks together. 
The proof is short so I'll leave it to the reader. 
Hint: 2*(n-1)! is the number of ways to arrange n people so that exactly one couple sticks together. 


I recommend trying small values of n to see examples of why this formula works. 
Here's a small bit of scratch work when n = 3
The double star notation indicates when A,B are together
<pre>
ABC **
ACB
BAC **
BCA
CAB **
CBA **
</pre>
There are 4 cases when A,B are together out of 6 permutations.
4/6 = 2/3 is the probability of A,B being together in either order.
This matches with the formula 2/n where n = 3. So this example helps confirm the formula works. I recommend trying it with n = 4 to see what you find. 


A similar question is found here
<a href="https://math.stackexchange.com/questions/114367/probability-people-sitting-in-a-row-linear-arrangement">https://math.stackexchange.com/questions/114367/probability-people-sitting-in-a-row-linear-arrangement</a>
The "solution" 1/63 mentioned on that page is NOT correct. I'm not sure why the professor made such a strange error. Instead the solution is 1/5 as the other people point out.


------------------------------------------------------------


Explanation for Part (b)


Let M and W represent the block of men and block of women respectively.
We could have MW or WM
So there are 2 ways to arrange these blocks.


Within any given such arrangement, we have 7! ways to arrange the 7 women within their respective block, and 5! ways to arrange the 5 men in their respective block.


2*7!*5! is the number of ways to arrange the people so that the genders are fully separated.
This is out of 12! ways to arrange all 12 people regardless if the genders are blocked together or not.


Divide the items mentioned.
(2*7!*5!)/(12!)
(2*7!*5!)/(12*11*10*9*8*7!)
(2*5!)/(12*11*10*9*8)
(2*5*4*3*2*1)/(12*11*10*9*8)
(4*3*2*1)/(6*11*2*9*8)
(2*1)/(6*11*2*3*2)
(1)/(6*11*2*3)
1/(66*6)
<font color=red>1/396</font> is the answer to part (b)
The steps above hopefully show how I cancelled things. Other routes are possible.
You can use a calculator to make quick work of this, but it still might be useful to know how to do this by hand. 
</font>