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Question Number 28805 by NECx last updated on 30/Jan/18

In a competition, a school awarded medals in different categories. 36 medals in dance,12 in dramatics and 18 medals in music.If these medals went to total 45,and only 4 persons got medals in all three catogories.Using set notations, how many received in exactly two of these categories?

$${In}\:{a}\:{competition},\:{a}\:{school}\:{awarded} \\ $$$${medals}\:{in}\:{different}\:{categories}. \\ $$$$\mathrm{36}\:{medals}\:{in}\:{dance},\mathrm{12}\:{in}\:{dramatics} \\ $$$${and}\:\mathrm{18}\:{medals}\:{in}\:{music}.{If}\:{these} \\ $$$${medals}\:{went}\:{to}\:{total}\:\mathrm{45},{and}\:{only} \\ $$$$\mathrm{4}\:{persons}\:{got}\:{medals}\:{in}\:{all}\:{three} \\ $$$${catogories}.{Using}\:{set}\:{notations}, \\ $$$${how}\:{many}\:{received}\:{in}\:{exactly} \\ $$$${two}\:{of}\:{these}\:{categories}? \\ $$

Answered by Rasheed.Sindhi last updated on 30/Jan/18

A:Set of medals in dance B:Set of medals in drama C:Set of medals in music n(A)=36 , n(B)=12 , n(C)=18 n(A∪B∪C)=45 , n(A∩B∩C)=4 −.−.−.−.−.− n(A∪B∪C)=n(A)+n(B)+n(C) −n(A∩B)−n(B∩C)−n(A∩C) +n(A∩B∩C) n(A∩B)+n(B∩C)+n(A∩C)= n(A)+n(B)+n(C)−n(A∪B∪C) +n(A∩B∩C) n(A∩B)+n(B∩C)+n(A∩C)=36+12+18−45+4 =25 Each of (A∩B) , (B∩C) & (A∩C) contains (A∩B∩C) So, Number of medals received in exactly two catagaries: n(A∩B)+n(B∩C)+n(A∩C)−3×n(A∩B∩C) =25−3×4=13

$$\mathrm{A}:\mathrm{Set}\:\mathrm{of}\:\mathrm{medals}\:\mathrm{in}\:\mathrm{dance} \\ $$$$\mathrm{B}:\mathrm{Set}\:\mathrm{of}\:\mathrm{medals}\:\mathrm{in}\:\mathrm{drama} \\ $$$$\mathrm{C}:\mathrm{Set}\:\mathrm{of}\:\mathrm{medals}\:\mathrm{in}\:\mathrm{music} \\ $$$$\mathrm{n}\left(\mathrm{A}\right)=\mathrm{36}\:,\:\mathrm{n}\left(\mathrm{B}\right)=\mathrm{12}\:,\:\:\mathrm{n}\left(\mathrm{C}\right)=\mathrm{18} \\ $$$$\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\cup\mathrm{C}\right)=\mathrm{45}\:,\:\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right)=\mathrm{4} \\ $$$$−.−.−.−.−.− \\ $$$$\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\cup\mathrm{C}\right)=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)+\mathrm{n}\left(\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right)−\mathrm{n}\left(\mathrm{B}\cap\mathrm{C}\right)−\mathrm{n}\left(\mathrm{A}\cap\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\:\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right)+\mathrm{n}\left(\mathrm{B}\cap\mathrm{C}\right)+\mathrm{n}\left(\mathrm{A}\cap\mathrm{C}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)+\mathrm{n}\left(\mathrm{C}\right)−\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right)+\mathrm{n}\left(\mathrm{B}\cap\mathrm{C}\right)+\mathrm{n}\left(\mathrm{A}\cap\mathrm{C}\right)=\mathrm{36}+\mathrm{12}+\mathrm{18}−\mathrm{45}+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{25} \\ $$$$\mathrm{Each}\:\mathrm{of}\:\left(\mathrm{A}\cap\mathrm{B}\right)\:,\:\left(\mathrm{B}\cap\mathrm{C}\right)\:\&\:\left(\mathrm{A}\cap\mathrm{C}\right)\:\mathrm{contains} \\ $$$$\:\:\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\mathrm{So}, \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{medals}\:\mathrm{received}\:\mathrm{in}\:\mathrm{exactly}\:\mathrm{two} \\ $$$$\mathrm{catagaries}: \\ $$$$\:\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right)+\mathrm{n}\left(\mathrm{B}\cap\mathrm{C}\right)+\mathrm{n}\left(\mathrm{A}\cap\mathrm{C}\right)−\mathrm{3}×\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{25}−\mathrm{3}×\mathrm{4}=\mathrm{13} \\ $$$$ \\ $$

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