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Which of the following shall be a compound proposition involving the propositions p, q and r, that is true when exactly two of the p, q and r are true and is false otherwise?

  1. $(p \vee q \wedge \rceil r) \vee (p \wedge q \wedge r) \wedge (\rceil p \wedge q \vee r)$
  2. $(p \wedge q \vee r) \wedge (p \wedge q \wedge r) \vee (\rceil q \wedge \rceil q \wedge \rceil r)$
  3. $(p \wedge q \wedge \rceil r) \vee (p \wedge \rceil q \wedge r) \vee (\rceil p \wedge q \wedge r)$
  4. $(p \vee r \wedge q) \vee (p \wedge q \wedge r) \vee (\rceil p \wedge q \wedge r)$

2 Answers

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(P⋂Q⋂R')    -> P and Q are true and R is False [exactly two are true]

U

(P⋂Q'⋂R)     -> P and R are true and Q is False [exactly two are true]

U

(P'⋂Q⋂R)     -> Q and R are true and P is False [exactly two are true]

edited by
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Catching the condition given in the question that "exactly two of the p, q and r are true and is false otherwise." 

by solving option (b) like   (1 1 1) ^ (1 1 1) v (0 0 0) = (1 1 1) also gives true but,

Option (c)  is correct because it matches the condition given in the question having exactly two of p,q and r are true.

Hence, option (c)  is correct.

Answer:

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