If there are $10$ stations on a route and the train has to be stopped at $4$ of them then the number of ways in which train can be stopped so that at least two stopping stations are consecutive is :
$ (a) \: \; {}_{7}C_{4} \qquad (b) \: \; {}_{10}C_{4} - {}_8C_{3} \qquad (c) \: \; {}_{10}C_{4}- {}_{7}C_{3} \qquad (d) \: \; {}_{8}C_{3}$
Number of ways such that the train did not stop at consecutive stations = $^7C_4$.
We can treat a particular stop as "S" and a non-stop as "D". Then simply, what we can do is arrange all the "D"s first, in this case, 6 D's. i.e _D_D_D_D_D_D_.
Notice there are 7 blanks (denoted by _) where we can insert our "S" into. This result will produce no consecutive stops.
Without restrictions, there will be $^{10}C_4$ ways to "arrange" the stops and non-stops. Therefore, number of ways such that the train stopped at at least 2 stations consecutively = $^{10}C_4$ - $^7C_4$.
Yet, I understand that it may not be one of the solution you have above. However, do take note that $^7C_4$ = $^7C_3$, so the answer (c) still stands.
Hint 1 There are ${10\choose 4}$ ways to choose differnt stoppings
Hint 2 If we stop train, the route will be divided into 5 parts, where the length of each part is counted as the number of stations in it. If there are no consecutive stops, then 3 inner parts must me longer than 0. Let
$$A=\{(a_1,a_2,a_3,a_4, a_5)|a_i \geq 0, a_1\cdot a_2\cdot a_3 \neq 0, a_i \in \mathbb{Z}, \sum a_i=(10-4)\}$$
$|A|={(10-4-3)+5-1 \choose (10-4-3)} = {7\choose 3}$