SOLUTION: Three friends board an airliner just before departure time. There are only 12 seats left, 4 of which are aisle seats. How many ways can the 3 people arrange themselves in available

When you see the words "at least one", more than likely the problem 
is best solved by first finding the number of arrangements which 
either DO or DO NOT have the restriction.  Then we calculate the 
number of ways that DO NOT have the restriction and subtract that 
from the first number. That leaves the number that DO have the 
restriction.  The restriction here is that at least one of the three
must sit in an aisle seat.

So first we calculate the number of ways the three can sit in any 
of the 12 seats, whether an aisle seat or not.

Suppose the three people are Tom, Dick, and Harry.  We can seat Tom 
in any of the 12 seats, Dick in any of remaining 11, and Harry in
any of the remaining 10.  That's 12×11×10 ways to seat the three 
anywhere, aisle seats or not. 

From that we must subtract the number of ways none of them sit in
aisle seats.   

There are 4 aisle seats and therefore 12-4 or 8 non-aisle seats.

We can seat Tom in a non-aisle seat in any of 8 ways, Dick in any of
the remaining 7 non-aisle seats, and Harry in any of the  remaining 6
non-aisle seats. That's 8×7×6 ways to seat the three in non-aisle 
seats.

Now we subtract the 8×7×6 from the 12×11×10:

12×11×10 - 8×7×6 = 1320 - 336 = 984.

Edwin