First look at how many ways you can arrange 3 trains on 3 tracks. There are 6 combinations: ABC, ACB, BAC, BCA, CAB, CBA. Now, how many ways can you choose 3 trains out of 5? The general formula for combinations is:
Combinations of t chosen from n = C(n,t) = n!/(t!x(n-t)!) where n! = n x (n-1) x (n-2) x...x 1.
Choosing 3 out of 5 = 5!/(3!2!) = (5x4x3x2x1)/((3x2x1)x(2x1)) = (5x4)/(2x1) = 10
So there are 10 ways to choose the 3 trains and 6 ways to arrange them on the 3 tracks. You have 60 combinations (10x6) once you have built the 5 trains.
You have 35 cars (5 trains of 7 cars each). How many ways can you select 7 cars for the first train?
Choosing 7 out of 35 = 35!/(7!28!) = (35x34x33x32x31x30x29)/(7x6x5x4x3x2x1) = over 6 million.
Essentially you have a lotto drawing for the cars and the combinations are HUGE. Note that this is only the first train. For the second train, you are choosing 7 cars out of the remaining 28, then choosing 7 out of 21 for the 3rd train, etc. All of these combinations get multiplied together just like the trains and tracks did above.
You'll certainly never run out of different trains to run, that's for sure! 
