1/a+1/b+1/c=3/4. If a,b and c are natural numbers, how many such triplets (a,b,c) exist?

Let us try to enumerate all the cases and count the number of triplets. 

Without loss of generality, we can assume that a <= b <= c.

Therefore, the maximum value 'a' can take is 4.
In the case where a = 2, 1/b + 1/c = 1/4 or c = 4*b/(4-b)

The only possible natural number solutions to this are b = 5, c = 20; b = 6, c= 12 and b = c = 8
In the case where a = 3, 1/b + 1/c = 5/12 or c = 12*b/(5b-12)

The only possible natural number solutions to this are b = 3, c = 12 and b = 4, c= 6
In the case where a = 4, 1/b + 1/c = 1/2 and the only possible case is a=b=c=4

So, let us write all the possible triplets again (2,5,20); (2,6,12); (2,8,8); (3,3,12); (3,4,6) and (4,4,4)

Hence, the number of triplets possible is 6 + 6 + 3 + 3 + 6 + 1 = 25

25

Hi,
Can you please elaborate the question? Is it 1 X b or 1/b and 1 X c or 1/c. Are their any conditions on a,b,c such as they are intergers, natural numbers etc?