How many options of a train if it is built from $5$ locomotives and $17$ wagons? 
John has a train building set containing $5$ locomotives and $17$ wagons.
a) How many possibilities of a train can John build if a single train
  needs 1 locomotive and 10 wagons to be built?
b) By adding a locomotive at the end, how many possibilites are there
  now?

a)
I assume I use the combinations formula here $\frac{n!}{r!(n-r)!}$
I then calculated the possibilites of the locomotives $\frac{5!}{1!(5-1)!}$ which gives me the result of $5$. Then I calculate the possibilites of the wagons $\frac{17!}{10!(17-10)!}$ with the result being $19448$.
After that I simply multiply the both possibilites to get the final solution $97240*4=399960$
b)
I would simply take the final solution from a) and multiply it by $4$ which equals to $388960$. 
My question is: Are these valid solutions?
 A: 
How many ways can John select one locomotive and ten wagons with which to build a train? 

John can select one of the five locomotives and ten of the seventeen wagons in 
$$\binom{5}{1}\binom{17}{10} = 997240$$
ways.
There are mistakes in your calculations.  You omitted a $9$ from $997240$.  It is not clear why you multiplied by $4$.    

How many ways can John select two locomotives and ten wagons with which to build a train?

John can two of the five locomotives and ten of the seventeen wagons in
$$\binom{5}{2}\binom{17}{10} = 1994480$$
ways.

In how many ways can John assemble a train in which one locomotive pulls ten wagons if the cars are distinguishable?

John must select one of the five locomotives and ten of the seventeen wagons, then arrange the ten wagons, which can be done in 
$$\binom{5}{1}\binom{17}{10}10!$$
ways.

In how many ways can John assemble a train with two locomotives and ten wagons if the locomotives are placed at the ends of the train and the cars are distinguishable?

John must select two of the five locomotives and ten of the seventeen wagons, arrange the two locomotives at the ends of the train, and arrange the ten wagons between them, which can be done in 
$$\binom{5}{2}\binom{17}{10}2!10!$$
ways.
A: No. $\frac{17!}{7! ×10!}$ will give you an integer. Also check your calculations. Use permutations formula and recalculate it.
