The number of ordered triplets of the truth values of $p, q$ and $r$ such that the truth value of the statement $(p \vee q) \wedge(p \vee r) \Rightarrow(q \vee r)$ is True, is equal to

Consider the following statements :
$P$ : Suman is brilliant
$Q$ : Suman is rich.
$R$ : Suman is honest
the negation of the statement

"Suman is brilliant and dishonest if and only if suman is rich" can be equivalently expressed as

The maximum number of compound propositions, out of $p \vee r \vee s , p \vee P \vee \sim s , p \vee \sim q \vee s$,

$\sim p \vee \sim r \vee s , \sim p \vee \sim r \vee \sim s , \sim p \vee q \vee \sim s$, $q \vee r \vee \sim s , q \vee \sim r \vee \sim s , \sim p \vee \sim q \vee \sim s$

that can be made simultaneously true by an assignment of the truth values to $p , q , r$ and $s$, is equal to

The converse of the statement $((\sim p) \wedge q) \Rightarrow r$ is

Negation of $p \wedge (\sim q \vee \sim r)$ is -

Let $p$ and $q$ be two Statements. Amongst the following, the Statement that is equivalent to $p \to q$ is