The number of choices for $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$, such that $$(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$$ is a tautology, is

We need to find how many choices for $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$ make $$(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$$ a tautology.

Recall that $$P \Rightarrow Q$$ is false only when $$P$$ is true and $$Q$$ is false. So the expression is a tautology if whenever $$(p \Delta q)$$ is true, at least one of $$(p \Delta \sim q)$$ or $$(\sim p \Delta q)$$ is also true.

Case 1: $$\Delta = \wedge$$ (AND)

$$(p \wedge q) \Rightarrow ((p \wedge \sim q) \vee (\sim p \wedge q))$$

When $$p = T, q = T$$: LHS = T. RHS = $$(T \wedge F) \vee (F \wedge T) = F \vee F = F$$.

So the implication is F. Not a tautology.

Case 2: $$\Delta = \vee$$ (OR)

$$(p \vee q) \Rightarrow ((p \vee \sim q) \vee (\sim p \vee q))$$

RHS: $$(p \vee \sim q) \vee (\sim p \vee q) = (p \vee \sim p) \vee (q \vee \sim q) = T \vee T = T$$.

Since RHS is always T, the implication is always T. Tautology.

Case 3: $$\Delta = \Rightarrow$$ (Implication)

$$(p \Rightarrow q) \Rightarrow ((p \Rightarrow \sim q) \vee (\sim p \Rightarrow q))$$

$$(\sim p \Rightarrow q)$$ is equivalent to $$(p \vee q)$$. This is false only when $$p = F, q = F$$.

When $$p = F, q = F$$: LHS = $$(F \Rightarrow F) = T$$. RHS = $$(F \Rightarrow T) \vee (T \Rightarrow F) = T \vee F = T$$.

When $$p = T, q = T$$: LHS = T. RHS = $$(T \Rightarrow F) \vee (F \Rightarrow T) = F \vee T = T$$.

When $$p = T, q = F$$: LHS = F. So implication is T.

When $$p = F, q = T$$: LHS = T. RHS = $$(F \Rightarrow F) \vee (T \Rightarrow T) = T \vee T = T$$.

All cases give T. Tautology.

Case 4: $$\Delta = \Leftrightarrow$$ (Biconditional)

$$(p \Leftrightarrow q) \Rightarrow ((p \Leftrightarrow \sim q) \vee (\sim p \Leftrightarrow q))$$

Note: $$(p \Leftrightarrow \sim q) = \sim(p \Leftrightarrow q)$$ and $$(\sim p \Leftrightarrow q) = \sim(p \Leftrightarrow q)$$.

So RHS = $$\sim(p \Leftrightarrow q) \vee \sim(p \Leftrightarrow q) = \sim(p \Leftrightarrow q)$$.

The expression becomes: $$(p \Leftrightarrow q) \Rightarrow \sim(p \Leftrightarrow q)$$.

When $$p = T, q = T$$: $$(T) \Rightarrow (F) = F$$. Not a tautology.

Therefore, exactly 2 choices ($$\vee$$ and $$\Rightarrow$$) make the expression a tautology.

The answer is Option B: 2.