Dear student,
The required number of ways=8P3
Use permutation
						

							No.of ways to select 3 stations among 8 stations = 8C3No.of possibilities of 3stations might be consecutive = 6No.Of possibilities of 2 stations among those 3 might be consecutive = 30So, resultant answer is 8C3 - 6 - 30 = 20
						

							
Total No of way can a train be stopped= 8c3=42
let those eight stations be {1 2 3 4 5 6 7 8}
{(1,2)(2,3)(3,4)(4,5)(5,6)(6,7)(7,8)}7  {(1,2,3)(2,3,4)(3,4,5)(4,5,6)(5,6,7)(6,7,8)}No.Of waysof 2 stations among those 3 might be consecutive =6No.of ways of 3stations might be consecutive = 
therefore Ans is 42-(6+7)=20 
						

							
The number of ways of selecting any 3 stations among the 8 stations = 8C3=56
The number of ways of selecting 3 stations in which all 3 stations are consecutive = 6
The number of ways of selecting 3 stations in which only 2 are consecutive = 2*5 + 5*4=30
Therefore the number of ways a train can be stopped at 3 stations such that no two of them are consecutive=56-6-30=20