For any element xi present in X,4 cases arises while making subsets Y and Z. Case- 1:xi∈Y,xi∈Z⇒Y∩Z≠φ Case- 2:xi∈Y,xi∉Z⇒Y∩Z=φ Case- 3:xi∉Y,xi∈Z⇒Y∩Z=φ Case- 4:xi∉Y,xi∉Z⇒Y∩Z=φ ∴ For every element, number of ways =3 for which Y∩Z=φ ⇒ Total ways =3×3×3×3×3[∵ no. of elements in set X=5] =35