(c) Let there be 10 positions of the car, where they can be parked i.e., 1, 2 ,..... 10 Now, two have exactly 5 cars between the two particular cars their position must be at 1 and 7, 2 and 8, 3 and 9 and 4 and 10. Now, these two cars can be arranged in 4 x 2 = 8 ways as their positions can change. Remaining 8 cars can be arranged in = 8! ways Total number of arrangement = 10! ways ∴ Required probability =