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+11 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

  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 (68.8k points)
edited by | 844 views

4 Answers

+22 votes
Best answer

Answer D

$\quad(P\cap Q\cap R)\cup (P^{c}\cap Q\cap R)\cup Q^{c}\cup R^{c}$

$=(P\cup P^{c})\cap (Q\cap R)\cup Q^{c}\cup R^{c}$

$=(Q\cap R)\cup Q^{c}\cup R^{c}$

$=(Q\cap R)\cup (Q\cap R)^{C}$

$= U.$

answered by Veteran (10.9k points)
edited by
+12 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 (3.2k points)
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 

i above link's Ques 

in I,
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 (2.1k points)
+3 votes

hope it might help....

answered by Veteran (22.9k points)

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