explain the solution

[akpareek]

How many ways two numbers can be chosen from number {1,2, 3, ......... 30}, such that sum of numbers is divisible by 3?

A) 90 B) 100 C) 190 D) 225 E) 435

[iamseer]

To find ways such that a+b is divisible by 3 where a,b:{1,2,3,...30}

Now a number when divided by 3 leaves a remainder of either a 0 or a 1 or a 2 (3n or 3n+1 or 3n+2 where n = 0 to 10)

for remainder 0: 3n
a can be selected in 10 ways and then b can be selected in 9 ways
i.e. if a=6; b can be 3, 9, 12, ...multiples of 3 <=30
- 10*9=90 ways

for remainder 1: 3n+1
a can be selected in 10 ways and then b has to be selected from the remaining to digits of type 3n+2
- 10*10=100 ways

for remainder 2: 3n+2:
a can be selected in 10 ways and then b can be selected in 10 ways from the remaining digits of type 3n+1
but we have already counted all these in case 2.

so total 90+100=190 ways

Thanks for sharing this one... took some time to get hold of it but enjoyed it nonetheless
cheers!!