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+18 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 (52.1k points)
edited by | 1.7k views

5 Answers

+25 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 Loyal (5.6k points)
edited by
+15 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 Active (3.1k 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 

+11 votes

so option d 

answered by Active (1.8k points)
this explanation made it so easy. thanks.....
+6 votes

hope it might help....

answered by Boss (40.9k points)
0 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.

answered by Junior (855 points)

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