Considering the equation x1x2x3=60=(2)2. (3) 1. (5)
Let xi=2αi,3βi,4γi,αi,βi,γi≥0 ⇒(2)αi+α2+α3⋅(3)βi+β2+β3⋅(5)γ1+γ2+γ3⋅(2)2⋅(3)1⋅(5)1 ⇒α1+α2+α3=2;β1+β2+β3=1;γ1+γ2+γ3=1
Total number of positive integral solutions = 4C2⋅3C2⋅3C2=6×3×3=54