Given, xyz=24 ⇒xyz=23⋅31 Let x=2a1⋅3b1, y=2a2⋅3b2, z=2a3⋅3b3 where, a1,a2,a3∈{0,1,2,3} b1,b2,b3∈{0,1} Case I a1+a2+a3=3 ∴ Non-negative solution =3+3−1C3−1=5C2=10 Case II b1+b2+b3=1 ∴ Non-negative solution =1+3−1C3−1=3C2=3 ∴ Total solutions =10×3=30