How many ordered pairs of positive integers (a, b) satisfy 1/a + 1/b = 1/6?

Clear the fractions: $6b+6a=ab$, so $ab-6a-6b=0$, and adding 36 to both sides factorises it as $(a-6)(b-6)=36$. Every divisor $d$ of 36 now gives a solution $a=6+d,b=6+36/d$, and $36=2^2\times 3^2$ has $(2+1)(2+1)=9$ divisors. The nine ordered pairs are (7, 42), (8, 24), (9, 18), (10, 15), (12, 12), (15, 10), (18, 9), (24, 8) and (42, 7). The trap is 5 — that is the count when order is ignored, since four of the pairs come in swapped couples and (12, 12) stands alone.