If-A-B-C-are-finite-sets-whose-elements-are-from-the-same-universal-set-U-and-n-A-denotes-the-number-of-element-in-the-set-A-a-Show-by-means-of-venn-diagram-that-n-A-B-n-A-n-B-n-A-B-

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Question Number 203545 by Mastermind last updated on 21/Jan/24

$$\mathrm{If}\mathrm{A},\mathrm{B},\mathrm{C}\mathrm{are}\mathrm{finite}\mathrm{sets}\mathrm{whose}\mathrm{elements}\mathrm{are}$$$$\mathrm{from}\mathrm{the}\mathrm{same}\mathrm{universal}\mathrm{set}\mathrm{U}\mathrm{and}\mathrm{n}\left(\mathrm{A}\right)$$$$\mathrm{denotes}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{element}\mathrm{in}\mathrm{the}\mathrm{set}\mathrm{A}$$$$\left(\mathrm{a}\right)\mathrm{Show}\mathrm{by}\mathrm{means}\mathrm{of}\mathrm{venn}\mathrm{diagram}\mathrm{that}$$$$\mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})$$$$\left(\mathrm{b}\right)\mathrm{Using}\mathrm{the}\mathrm{fact}\mathrm{that}(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})=(\mathrm{A}\cup \mathrm{B})\cup \mathrm{C}$$$$=\mathrm{A}\cup (\mathrm{B}\cup \mathrm{C})\mathrm{deduce}\mathrm{an}\mathrm{expression}\mathrm{for}$$$$(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})$$$$\left(\mathrm{c}\right)\mathrm{If}\mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}(\mathrm{A}\cap \mathrm{B}),\mathrm{what}\mathrm{can}\mathrm{be}\mathrm{said}$$$$\mathrm{about}\mathrm{A}\mathrm{and}\mathrm{B}?\mathrm{How}\mathrm{did}\mathrm{you}\mathrm{reach}\mathrm{your}\mathrm{conclusion}.$$$$$$$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{in}\mathrm{advance}$$

Answered by AST last updated on 21/Jan/24

$$b)n\mid (A\cup B)\cup C\mid =n[P\cup C]whereP=A\cup B$$$$n(P\cup C)=n\left(P\right)+n\left(C\right)-n(P\cap C)$$$$=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(P\cap C)$$$$P\cap C=(A\cap B)\cap C=(A\cap C)\cup (B\cap C)$$$$n\left(PnC\right)=n(A\cap C)+n(B\cap C)-n[(A\cap C)\cap (B\cap C)]$$$$\Rightarrow n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+\left(C\right)-n(A\cap B)$$$$-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$$$$$$$$c)Supposeoneofthesets,sayA,isnotasubset$$$$oftheother,B.Then,\mathrm{\exists }{a}_1\in As.t.a_1\notin B$$$$\Rightarrow n(A\cup B)>n(A\cap B)\Rightarrow A\subseteq B\Rightarrow n(A\cap B)=n\left(A\right)$$$$SupposeB\mathrm{\setminus }A\ne \mathrm{\varnothing }\Rightarrow n\left(B\mathrm{\setminus }A\right)=k⩾1$$$$thenn(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)=n\left(B\right)$$$$=n\left(B\mathrm{\setminus }A\right)+n\left(A\right)=k+n\left(A\right)>n(A\cap B)$$$$\Rightarrow n\left(B\mathrm{\setminus }A\right)=\mathrm{\varnothing }\Rightarrow A=B$$

Commented by Mastermind last updated on 21/Jan/24

$$\mathrm{Thank}\mathrm{you}\mathrm{but}\mathrm{whats}\mathrm{the}\mathrm{meaning}\mathrm{of}\mathrm{s}.\mathrm{t}.?$$

Commented by Mastermind last updated on 21/Jan/24

$$\mathrm{besides}\mathrm{i}\mathrm{could}\mathrm{not}\mathrm{identify}\mathrm{solution}\mathrm{a}\mathrm{and}\mathrm{solution}\mathrm{b}$$

Commented by AST last updated on 21/Jan/24

$$suchthat$$

Commented by Mastermind last updated on 21/Jan/24

$$\mathrm{Thank}\mathrm{you}\mathrm{but}\mathrm{help}\mathrm{me}\mathrm{separate}\mathrm{the}\mathrm{solution}$$$$\mathrm{separately}.$$$$$$$$\mathrm{Thank}\mathrm{you}$$

Commented by Mastermind last updated on 21/Jan/24

$$\mathrm{Where}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{A}?$$

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