maths.... permutation and combination...?

Let
A = { x ∣ x = 3k , for some integer k } = { 3 , 6 , 9 , ..... , 30 }
B = { x ∣ x = 3k + 1 , for some integer k } = { 1 , 4 , 7 , ..... , 28 }
C = { x ∣ x = 3k + 2 , for some integer k } = { 2 , 5 , 8 , ..... , 29 }
Then, n(A) = n(B) = n(C) = 10

There are only 2 cases for the sum is divisible by 3 :
Case ( i )
Select 2 numbers from set A
C( 10 , 2 ) = 10*9 / (2*1) = 45

Case ( ii )
Select a number from set B, and select a number from C
10*10 = 100

45 + 100 = 145 ... Ans