40 40 votes If $P, Q, R$ are subsets of the universal set U, then $$(P\cap Q\cap R) \cup (P^c \cap Q \cap R) \cup Q^c \cup R^c$$ is$Q^c \cup R^c$$P \cup Q^c \cup R^c$$P^c \cup Q^c \cup R^c$$\cup$ Set Theory & Algebra gatecse-2008 normal set-theory&algebra set-theory + – Kathleen 15.4k views answer comment Share Follow Print See 1 comment 1 1 comment reply js__ commented Oct 30, 2025 reply Follow flag convert to Boolean expr then do 3 3 replyShare Please log in or register to add a comment.
Best answer 58 58 votes Answer D $\quad(P\cap Q\cap R)\cup (P^{c}\cap Q\cap R)\cup Q^{c}\cup R^{c}$ $=(P\cup P^{c})\cap (Q\cap R)\cup Q^{c}\cup R^{c}$ $=(Q\cap R)\cup Q^{c}\cup R^{c}$ $=(Q\cap R)\cup (Q\cap R)^{C}$ $= U.$ Anu answered Jun 25, 2015 • edited Nov 26, 2017 by pavan singh Anu comment Share Follow 0 reply Please log in or register to add a comment.
49 49 votes so option d Çșȇ ʛấẗẻ answered Jan 5, 2017 Çșȇ ʛấẗẻ comment Share Follow See all 3 Comments 3 3 Comments reply skeltro commented Aug 28, 2018 reply Follow flag this explanation made it so easy. thanks..... 1 1 replyShare Hira Thakur commented Aug 27, 2023 reply Follow flag $\cap\equiv $AND gate $\cup\equiv $ OR gate 3 3 replyShare js__ commented Oct 30, 2025 reply Follow flag best method!! 0 0 replyShare Please log in or register to add a comment.
21 21 votes Can we treat these like Boolean expression and solve? Like PQR + P'QR + Q' + R'. and minimise this. Is this method always correct? @Praveen Sir? @Arjun Sir? Aspi R Osa answered Jan 7, 2016 Aspi R Osa comment Share Follow See all 7 Comments 7 7 Comments reply Show 4 previous comments Praveen Saini commented Aug 7, 2017 reply Follow flag $A-B = A \cap B'$ $P\Delta (Q\cap R)$= P-(Q.R) = P.(QR)' = PQ'+PR' that is $(P\Delta Q) \cup (P\Delta R)$ 2 2 replyShare Gate Ranker18 commented Aug 8, 2017 reply Follow flag Praveen Saini sir whats wrong in my explanation plz verify 0 0 replyShare Rajsukh Mohanty commented Dec 18, 2023 reply Follow flag @Praveen Saini sir, i think the $\Delta$ notation represents symmetric difference. but, you have used it here in $P \Delta (Q \cap R)$ as set difference. it should be, i think $P - (Q \cap R)$. 0 0 replyShare Please log in or register to add a comment.
21 21 votes hope it might help.... akash.dinkar12 answered Jul 22, 2017 akash.dinkar12 comment Share Follow See all 3 Comments 3 3 Comments reply Doodle) commented Jan 8, 2020 reply Follow flag But the problem with this solution is " the diagram"! How did u come to the conclusion that the diagram looks like the one you have drawn ? They haven't said anything Abt their intersection right? All three can be independent sets and still be a subset of U! Do correct me if wrong:) 2 2 replyShare bencodes12 commented May 7, 2023 reply Follow flag IF THEIR IS NO INTERSECTION THE VALUE IN THE INTERSECTION BOX OF THE DIAGRAM WILL COME OUT TO BE ZERO AUTOMATICALLY. 3 3 replyShare Tushar Rana commented Apr 10 reply Follow flag @Doodle) It's because it cover's all the possibilities that the 3 sets can cover. 0 0 replyShare Please log in or register to add a comment.
5 5 votes Treating as a boolean expression like suggested in an answer here: PQR + P'QR + Q' + R' = (P+P') QR + Q' + R' = QR + Q' + R' = QR + Q'R' + Q'R + R' = R(Q+Q') + R'(Q'+1) = R + R' = 1 Also R+R' means RUR' so its equal to U. ShamikBanerjee answered Mar 9, 2019 ShamikBanerjee comment Share Follow 0 reply Please log in or register to add a comment.
5 5 votes If someone is good in digital Logic part or in vein Diagram part, than this question is easy for them :) Just convert Union into + and intersection in . and try to solve it. else Vein diagram becomes very easy for understanding. If Understood UpVoted :) Surya_Dev Chaturvedi answered Jan 15, 2021 Surya_Dev Chaturvedi comment Share Follow 0 reply Please log in or register to add a comment.