In how many ways can 8 people be seated in a row? What is the probability that Al sits next to Bob? For the first part I got 8!, which seems logical enough, but my attempt at the second part was a disaster. 
I decided that there were $C_{7,1}C_{2,1}$ ways for Al and Bob to sit together and then $6!$ ways for the other 6 people to sit, all over 8! total arrangements, which got me $$\frac{C_{7,1}C_{2,1}6!}{8!}=180$$ which is not even a probability. How would I approach this problem, and what value am I supposed to get?
 A: Try gluing Al and Bob together and then count the arrangements of the now  seven distinct entities.
Then remember that Al and Bob can turn around in place to reverse their order in the line... 
A: Although I use the same method as DJohnM,
your formula is also correct, you have just not computed properly to get the ans of 0.25 
A: Well, imagine Al and Bob are attached as single person.  There are 7! to seat these 7 sort of people.  But Al can be to the left or the right of Bob so there are 2*7! ways to do this.
So the probability is:  2*7!/8! = 1/4.  
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Another way to view it is that there are 8 places to seat Al leaving 7 places to seat Bob for 8x7 ways to seat Al and Bob.  Of these ways there are 7 which are nest to each other and Al can sit to the left or right of Bob so there are 14 ways they can be together.  That's 14/8x7 = 1/4 probability.
A: You should count the number of the ways 8 people can be seated so that Al is next to Bob and divide that by the number of all the ways 8 people can be seated, in order to get a probability that Al will be seated next to row, when you pick the order at random.
The second number (denominator) is of course $8!$. The nominator can be counted in many ways, one of which is: 
Al can be seated on 8 chairs. If he's seated on chair number 1 or 8, Bob must sit at 2 or 7, respectively. If Al's seated elsewhere, Bob can be seated at one of two chairs next to Al. In each of these cases, other 6 people can be seated in 6! ways. Therefore, there are $(2\cdot 1 + 6 \cdot 2)\cdot 6!$ ways 8 people can be seated so that Al is next to Bob.
A: You can think about it this way. 
There are 7 pairs of adjacent seats which you can put Al and Bob in. For each of those choices, there are two ways could sit, Al on the left Bob on the right or vice versa. Therefore there are 7*2=14 ways you could sit Al and Bob. I'm not sure what the C's are in your answer, but the factors of 6! and 8! are right. 
I then get an answer of 14*6!/8! = 1/4=25%.
