Question 1207172
<pre><b><font size=4>
Here is an alternate way to solve the problem using stars and bars.  

Let's suppose the correct interpretation for a successful arrangement is for one
of the pair A and B, to be in the far left chair, and the other to be in the far
right chair.

There are 2 ways to arrange A and B on the ends.

There are 6*5=30 ways to arrange C and D between them in the 6 chairs between
them.

So the number of SUCCESSFUL arrangements is 2*6*5=60.

Now let's look at the number of POSSIBLE arrangements. Take this sample

CDBA.  There are 4!=24 arrangements like this one.

We must now insert 4 empty chairs among them. 

This involves the number of partitions of 4 of length 5.

We make a row of 4 stars and 4 bars, like this sample:

**||*|*|

That is the partition 2+0+1+1+0=4.  That's because 
there are: 
2 stars left of the 1st bar, 
0 stars between the 1st and 2nd bar,
1 star between the 2nd and 3rd bar,
1 star between the 3rd and 4th bar, and
0 stars after the 4th bar.

This sample partition of 4 of length 5, which is 2+0+1+1+0 makes the sample 
ADCB to become:
 
_ _ A D _ C _ B

where the blanks represent the 4 empty chairs.

Now we find the number of ways to insert the 4 empty chairs among the 4 people.

There are 4 indistinguishable stars and 4 indistinguishable bars.

So there are {{{8!/(4!4!)}}}{{{""=""}}}{{{70}}} ways to put 4 empty chairs 

among the 4 people.

So let's put what we've said here all together: 

There are 4!=24 ways to form something like this: ADCB.

Then there are 70 ways to form something like this _ _ A D _ C _ B

That's 24x70 = 1680 possible ways.

So the probability is {{{60/1680}}}{{{""=""}}}{{{1/28}}}

Edwin</pre></b></font>