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**Answer: 3**

Let’s simplify the propositional function a bit:

$= \lnot p \leftrightarrow (q \land \lnot(p \to r))$

$= \lnot p \leftrightarrow (q \land \lnot (\lnot p \lor r))$

$= \lnot p \leftrightarrow (q \land p \land \lnot r)$

Now we know $a \leftrightarrow b$ is true when either both $a$ and $b$ are false or both $a$ and $b$ are true.

__Now considering case (i) when both $a$ and $b$ are false __

It is only possible when $p$ is true so $\lnot p$ is false, now to $(q \land p \land \lnot r)$ to be false we have following options

$q$ | $r$ | $(q∧p∧¬r)$ |

F | F | F |

F | T | F |

T | F | T |

T | T | F |

So $(q∧p∧¬r)$ is false only for 3 options.

__Now considering case (ii) when both $a$ and $b$ are true __

Here $a$ to be true $p$ must be false but if $p$ is false $(q∧p∧¬r)$ can’t be true.

So total combination of $p$, $q$ and $r$ for which $\lnot p \leftrightarrow (q \land \lnot(p \to r))$ is true is only 3.