There are 24 cars in the parking lot outside of my building. They are all either 2-door or 4-door, are either white, black, or red, and are

I assume you're sayng that all the cars differ in at least one aspect......so...there are 24 "different" cars, thusly:

(H, W,2)          (V, W,2)            (F, W, 2)          (C, W,2) 

(H, W,4)          (V, W,4)            (F, W, 4)          (C, W,4)                   

(H, R, 2)          (V, R,2)             (F,  R,  2)         (C, R, 2)

(H, R, 4)          (V, R 4)             (F, R,  4)          (C, R, 4)  

Notice that there are 4 ways  to  pair the  Hondas with a "different" Volvo, "different" Ford  or "different" Cadillac.  So, 4 + 4 + 4 = 12

And there are 4 ways  to pair the Volvos wth a "different" Ford or "different' Cadillac. {We've already paired each with a "different" Honda.} So 4 + 4 = 8

And since we've already paired the Fords with a "different" Honda or Volvo, we have 4 ways that we can pair them with a "different" Cadillac.

So there are 12 + 8 + 4 = 24 "different" pairings

And the total number of possible pairs made by selecting any 2 cars from 24  = C(24,2) = 276

So 24 / 276 = 2/23 = about an 8.7% chance of pairing two "different" cars......

hmm 2/23 didnt work for me

Since there are  possible combinations of the above properties, each possible type of car is represented exactly once. Once we choose the first car, the second car has  possibility for the doors to be different,  possibilities for the color to be different, and  possibilities for the brand to be different. Thus there are  cars that have no properties in common with the first car. We could have chosen this second car in 23 equally-likely ways, and hence our answer is 6/23.

So the answer is 6/23 :)

don't cheat on aops problems guys