Y⊆X,Z⊆X Let a∈X, then we have following chances that (1) a∈Y,a∈Z (2) a∈/Y,a∈Z (3) a∈Y,a∈/Z (4) a∈/Y,a∈/Z We require Y∩Z=ϕ Hence (2), (3), (4) are chances for 'a' to satisfy Y∩Z=ϕ. ∴Y∩Z=ϕ has 3 chances for a. Hence for five elements of X, the number of required chances is 3×3×3×3×3=35