Let x = number of apples,
y = number of oranges and
z = number of mangoes.
Here, (x + y + z) = 20
For favourable cases, we should have x < y < z.
The total number of non-negative integral sets of (x,y,z)
satisfying (x + y + z) = 20 are ²²C₂ = 231.
In these sets, there would be cases when two of the
three variables are equal. The equal variables could
be (0,0) or (1,1) or (2,2)… or (10,10). The other third
variable in each of these cases would be different.
Hence, each of these 11 cases would generate 3
unfavourable sets of (x,y,z). For example, (x,y,z) =
(0,0,20), (0,20,0), (20,0,0) are three unfavourable
cases, because we want x < y < z.
Hence, excluding these 33 cases, there are (231 – 33)
= 198 sets. In these sets, there would be equal number
of sets for each of (x < y < z), (y < z < x), (z < x < y),
(x < z < y), (y < x < z), (z < y < x). Hence, our favourable
number of cases would be
198/6= 33 and hence the
required probability =33/231=1/7