For any number $n \in \Bbb{Z}$, define
$$ f(n) = \Omega(n) - \sum_{q \mid n} \Omega(n/q) + \sum_{q,q' \mid n}\Omega(n/(qq')) - \dots \pm \Omega(1),$$
where each sum is taken over only prime divisors and includes multiple copies of the same prime (multiplicity).
Will this sum always be zero?
Is there a way to view it using inclusion-exclusion?
Let $n=q_1 q_2\dots q_r$ be a factorization into primes and consider the set of indices $Q=[r]=\{1,\dots,r\}$ so that $\Omega(n)=|Q|=r$.
For each $i\in[r]$ let $Q_i=Q\setminus \{i\}$ and note that $\Omega(n/q_i) = |Q_i|.$
When $r>1$, the inclusion-exclusion principle for $Q=Q_1\cup Q_2\cup\dots\cup Q_r$ yields $$ |Q|\ - \sum_{1\le i_1\le r} |Q_{i_1}|\ + \sum_{1\le i_1<i_2\le r} |Q_{i_1}\cap Q_{i_2}|\ - \ \cdots\ \pm |Q_1\cap Q_2\cap \dots \cap Q_r| = 0. $$ For $1\le i_1<\dots<i_n\le r$ we have $$ n/\prod_{k=1}^n q_{i_k} = \prod_{i\in Q'} q_i \qquad\text{for}\qquad Q'=Q\setminus\{i_1,\dots,i_n\} = Q_{i_1}\cap\dots\cap Q_{i_n}. $$ Hence, $$ \Omega\left(n/\prod_{k=1}^n q_{i_k}\right) = \left|Q_{i_1}\cap\dots\cap Q_{i_n}\right| $$ and the above identity is exactly $f(n)=0$.
When $r=1$, so that $n=p$ is a prime, we have $$ f(p) = \Omega(p) - \Omega(1) = 1 - 0 = 1 \neq 0. $$