8 persons are sitting in a row,from which 3 are selected at random.Find the probability that no two selected persons are together?

Let the 8 people be numbered 1 to 8.

Number of people left unselected before the first person is selected = a
Number of people left unselected between the first and second selected persons = b
Number of people left unselected between the second and third selected persons = c
Number of people left unselected after the third person is selected = d
For example, if people numbered 1, 4 and 7 are selected, a = 0, b = 2, c = 2 and d = 1
Number of selected people = 3, number of unselected people = 5
So, a + b + c + d = 5 where a, d >= 0 and b, c >= 1
=> a + b + c + d = 3 where a, b, c, d >= 0
= $$^{3 + 4 - 1}C_{4 - 1} = ^6C_3 = 20$$
Number of ways of selecting any three people = $$^8C_3$$ = 56
Required probability = 20/56 = 5/14