29 votes 29 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 $Q^c \cup R^c$ $P \cup Q^c \cup R^c$ $P^c \cup Q^c \cup R^c$ U Set Theory & Algebra gatecse-2008 normal set-theory&algebra set-theory + – Kathleen asked Sep 11, 2014 • edited Jun 21, 2017 by Silpa Kathleen 9.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 44 votes 44 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 See all 0 reply Please log in or register to add a comment.
37 votes 37 votes so option d Çșȇ ʛấẗẻ answered Jan 5, 2017 Çșȇ ʛấẗẻ comment Share Follow See all 2 Comments See all 2 2 Comments reply skeltro commented Aug 28, 2018 reply Follow Share this explanation made it so easy. thanks..... 1 votes 1 votes Hira Thakur commented Aug 27, 2023 reply Follow Share $\cap\equiv $AND gate $\cup\equiv $ OR gate 0 votes 0 votes Please log in or register to add a comment.
18 votes 18 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 See all 7 7 Comments reply Praveen Saini commented Jan 7, 2016 reply Follow Share Yes absolutely correct , will get 1 , that is U 19 votes 19 votes Gate Ranker18 commented Aug 6, 2017 reply Follow Share Praveen Saini if use Aspi R Osa 's method and found P.PQ then this equivalent to PQ or we take it as P.PQ ? 0 votes 0 votes Praveen Saini commented Aug 6, 2017 reply Follow Share Yes it will be PQ 0 votes 0 votes Gate Ranker18 commented Aug 7, 2017 reply Follow Share Praveen Saini sir https://gateoverflow.in/3562/gate2006-it-23 i above link's Ques in I, LHS=P+QR-PQR RHS=(P+Q-PQ).(P+R-PR) =P+PR-PR+PQ+QR-PQR-PQ-PQR+PQR =P+QR-PQR LHS=RHS So I is true but original ans is I is false plz verify 0 votes 0 votes Praveen Saini commented Aug 7, 2017 reply Follow Share $A-B = A \cap B'$ $P\Delta (Q\cap R)$= P-(Q.R) = P.(QR)' = PQ'+PR' that is $(P\Delta Q) \cup (P\Delta R)$ 2 votes 2 votes Gate Ranker18 commented Aug 8, 2017 reply Follow Share Praveen Saini sir whats wrong in my explanation plz verify 0 votes 0 votes Rajsukh Mohanty commented Dec 18, 2023 reply Follow Share @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 votes 0 votes Please log in or register to add a comment.
18 votes 18 votes hope it might help.... akash.dinkar12 answered Jul 22, 2017 akash.dinkar12 comment Share Follow See all 2 Comments See all 2 2 Comments reply Doodle) commented Jan 8, 2020 reply Follow Share 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:) 0 votes 0 votes bencodes12 commented May 7, 2023 reply Follow Share IF THEIR IS NO INTERSECTION THE VALUE IN THE INTERSECTION BOX OF THE DIAGRAM WILL COME OUT TO BE ZERO AUTOMATICALLY. 0 votes 0 votes Please log in or register to add a comment.