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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

  1. $Q^c \cup R^c$
  2. $P \cup Q^c \cup R^c$
  3. $P^c \cup Q^c \cup R^c$
  4. U
asked in Set Theory & Algebra by Veteran (66.1k points) 1148 2197 2522
edited by | 612 views

4 Answers

+18 votes
Best answer
Answer D

$(P∩Q∩R)\cup (P^c∩Q∩R)\cup Q^c\cup R^c

\\=(P∪P^c)∩(Q∩R)∪Q^c∪R^c

\\=(Q∩R)∪Q^c∪R^c

\\=(Q∩R)∪(Q∩R)^C

\\= U$
answered by Veteran (10.7k points) 15 62 101
selected by
+9 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?

answered by Loyal (3.1k points) 5 36 78
Yes absolutely correct , will get 1 , that is U

 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 ?

Yes it will be PQ

 Praveen Saini sir 

http://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

$A-B = A \cap B'$
$P\Delta (Q\cap R)$= P-(Q.R) = P.(QR)' = PQ'+PR' that is $(P\Delta Q) \cup (P\Delta R)$

 Praveen Saini  sir 

whats wrong in my explanation 

plz verify 

+8 votes

so option d 

answered by Active (1.5k points) 5 14 50
+1 vote

hope it might help....

answered by Veteran (17.3k points) 7 12 40


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