Question

Verify $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(c\right)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$ for the sets given below:
(i) A={4,5,6}, B={5,6,7,8} and C={6,7,8,9}
(ii) A={a, b, c, d, e}, B={x, y, z} and C={a, e, x}

Answer 1

(i) The given sets are $A=\{4,5,6\}$, $B=\{5,6,7,8\}$ and $C=\{6,7,8,9\}$

Now we find the following:

$A=\{4,5,6\}\Rightarrow n\left(A\right)=3B=\{5,6,7,8\}\Rightarrow n\left(B\right)=4C=\{6,7,8,9\}\Rightarrow n\left(C\right)=4$

$A\cap B=\{4,5,6\}\cap \{5,6,7,8\}=\{5,6\}\Rightarrow n(A\cap B)=2B\cap C=\{5,6,7,8\}\cap \{6,7,8,9\}=\{6,7,8\}\Rightarrow n(B\cap C)=3A\cap C=\{4,5,6\}\cap \{6,7,8,9\}=\left\{6\right\}\Rightarrow n(A\cap C)=1$

$A\cap B\cap C=\{4,5,6\}\cap \{5,6,7,8\}\cap \{6,7,8,9\}=\left\{6\right\}\Rightarrow n(A\cap B\cap C)=1A\cup B\cup C=\{4,5,6\}\cup \{5,6,7,8\}\cup \{6,7,8,9\}=\{4,5,6,7,8,9\}\Rightarrow n(A\cup B\cup C)=6$

Now consider

$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\Rightarrow 6=3+4+4-2-3-1+1\Rightarrow 6=3+4+4+1-(2+3+1)\Rightarrow 6=12-6\Rightarrow 6=6$

which is true.

Hence, $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

(ii) The given sets are $A=\{a,b,c,d,e\}$, $B=\{x,y,z\}$ and $C=\{a,e,x\}$

Now we find the following:

$A=\{a,b,c,d,e\}\Rightarrow n\left(A\right)=5B=\{x,y,z\}\Rightarrow n\left(B\right)=3C=\{a,e,x\}\Rightarrow n\left(C\right)=3$

$A\cap B=\{a,b,c,d,e\}\cap \{x,y,z\}=\left\{\varphi \right\}\Rightarrow n(A\cap B)=0B\cap C=\{x,y,z\}\cap \{a,e,x\}=\left\{x\right\}\Rightarrow n(B\cap C)=1A\cap C=\{a,b,c,d,e\}\cap \{a,e,x\}=\{a,e\}\Rightarrow n(A\cap C)=2$

$A\cap B\cap C=\{a,b,c,d,e\}\cap \{x,y,z\}\cap \{a,e,x\}=\left\{\varphi \right\}\Rightarrow n(A\cap B\cap C)=0A\cup B\cup C=\{a,b,c,d,e\}\cup \{x,y,z\}\cup \{a,e,x\}=\{a,b,c,d,e,x,y,z\}\Rightarrow n(A\cup B\cup C)=8$

Now consider

$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\Rightarrow 8=5+3+3-0-1-2+0\Rightarrow 8=5+3+3+0-(0+1+2)\Rightarrow 8=11-3\Rightarrow 8=8$

which is true.

Hence, $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$