$\begin{align} & a)\text{ If }n(A)\,\text{denotes the number of elements contained in the set }A,\text{ prove that } \\ & n(A)+n(B)=n(A\cup B)+n(A\cap B) \\ & (b)\text{Prove that } \\ & n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\\end{align}$
$\begin{align} & n(A)=x \\ & n(B)=y \\ & n(A\cap B)=z \\ & n(A\cap B')=x-z \\ & n(B\cap A)=y-z \\ & \therefore n(A\cup B)=n(A\cap B')+n(A\cap B)+n(B\cap A') \\ & n(A\cup B)=x-z+z+y-z \\ & n(A\cup B)=x+y-z \\ & n(A\cup B)=n(A)+n(B)-n(A\cap B) \\ & n(A)+n(B)=n(A\cup B)+n(A\cap B) \\ & \\ & \text{Let }n(A\cup B\cup C)=n(A\cup D) \\ & \text{where }B\cup C=D \\ & n(A\cup D)=n(A)+n(D)-n(A\cap D) \\ & n(A\cup D)=n(A)+n(B\cup C)-n(A\cap D) \\ & n(A\cup D)=n(A)+n(B)+n(C)-n(B\cap C)-n(A\cap D) \\ & n(A\cup D)=n(A)+n(B)+n(C)-n(B\cap C)-n[A\cap (B\cup C)] \\ & n(A\cup D)=n(A)+n(B)+n(C)-n(B\cap C)-\left[ n(A\cap B)+n(A\cap C)-n(A\cup B\cup C) \right] \\ & n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C) \\\end{align}$