Correct Answer - D
The number of ways of choosing two numbers from the given set is .3nC2.
Let us divide given 3n numbers into three groups G1,G2 and G3 as follows :
G1:3,6,9,..,3n
G2:1,4,7,10,..,3n-1
G3:2,5,8,11,..,3n-2
We have, a3+b3=(a+b)3-3ab(a+b)
Therefore, a3+b3 will be divisible by 3, if a+b is divisible by 3.
Now, a+b will be divisible by 3 in the following cases :
(i) Both the numbers belong to the first group.
(ii) One of the two numbers belongs to second group and one of them belongs to the third group.
∴ Favourable number of elementry events =.nC2+.nC1×.nC1
)
.3nC2
=
1
3