BITSAT 2012 Paper

We have : 30 = 2 × 3 × 5. So, 2 can be assigned to either a or b or c i.e. 2 can be assigned in 3 ways. Similarly, each of 3 and 5 can be assigned in 3 ways. Thus, the number of solutions is 3 × 3 × 3 = 27.