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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
edited | 1.7k views

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

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

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

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

$= U.$

edited

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?

+11
Yes absolutely correct , will get 1 , that is U
0

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
Yes it will be PQ
0

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

+2
$A-B = A \cap B'$
$P\Delta (Q\cap R)$= P-(Q.R) = P.(QR)' = PQ'+PR' that is $(P\Delta Q) \cup (P\Delta R)$
0

whats wrong in my explanation

plz verify so option d

0
this explanation made it so easy. thanks..... hope it might help....

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.