Question

Observe the figure 1.18 and verify the following equations.
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)

figure

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}n(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})=n\left(\mathrm{A}\right)+n\left(\mathrm{B}\right)+n\left(\mathrm{C}\right)-n(\mathrm{A}\cap \mathrm{B})-n(\mathrm{B}\cap \mathrm{C})-n(\mathrm{C}\cap \mathrm{A})+n(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}) \\ \mathrm{From}\mathrm{the}\mathrm{figure},\mathrm{we}\mathrm{have}: \\ n\left(A\cup B\cup C\right)=9...\left(1\right) \\ n\left(\mathrm{A}\right)=5 \\ n\left(\mathrm{B}\right)=5 \\ n\left(\mathrm{C}\right)=4 \\ n(\mathrm{A}\cap \mathrm{B})=2 \\ n(\mathrm{B}\cap \mathrm{C})=2 \\ n(\mathrm{C}\cap \mathrm{A})=2 \\ n(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C})=1 \\ \therefore n\left(\mathrm{A}\right)+n\left(\mathrm{B}\right)+n\left(\mathrm{C}\right)-n(\mathrm{A}\cap \mathrm{B})-n(\mathrm{B}\cap \mathrm{C})-n(\mathrm{C}\cap \mathrm{A})+n(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C})=5+5+4-2-2-2+1=9...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ n(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})=n\left(\mathrm{A}\right)+n\left(\mathrm{B}\right)+n\left(\mathrm{C}\right)-n(\mathrm{A}\cap \mathrm{B})-n(\mathrm{B}\cap \mathrm{C})-n(\mathrm{C}\cap \mathrm{A})+n(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}) \end{gathered}$$

Hence, proved.