The correct option is A 35
Let x1be the number of stations before the first halting station, x2 between first and second, x3 between second and third, x4 between third and fourth and x5 on the right of the fourth station, then
x1≥0, x2, x3, x4 ≥1 and x5≥0
such that x1+x2+x3+x4+x5=6
Now, the total number of ways is the number of solutions of the above equation.
Let y2≥0, y3≥0, y4≥0
∴ x1+x2+x3+x4+x5=6⇒ x1+(y2+1)+(y3+1)+(y4+1)+x5=6⇒ x1+y2+y3+y4+x5=3
∴ The number of solutions =3+5−1C5−1=7C4=35