Number of ways $x_{n+1} \lor (x_1 \land x_2 \land \cdots\land x_n)$ where $n \geq 2$ results in $F$ I'm trying to find the number of distinct assignments for each of the $x$ variables exist such that $x_{n+1} \lor (x_1 \land x_2 \land \cdots\land x_n)$ where $n \geq 2$ results in $F$
I know $x_{n+1}$ has to be $F,$ so that's one choice.
I know that $(x_1 \land x_2 \land \cdots\land x_n)$ must be $F$ which means at least one of the terms inside has to be F. So that's $1 \cdot 2^{n - 1}$ choices.
I'm getting the answer to be $2^{n - 1}$ choices but if I try $n = 2$ and draw a truth table I get $3$ choices (as opposed to 2 from my answer). If $n = 3,$ I get $7$ choices (as opposed to $4$ from my answer).
What am I doing wrong?

As you said, you need at least one of the $n$ propositions $x_k$ to be false. Thus, there is one such assigment for every non-empty subset of $\{x_1,\ldots,x_n\}$. This set has $2^n$ subsets, only one of which is empty, so it has $2^n-1$ non-empty subsets, not $2^{n-1}$, and there are therefore $2^n-1$ assigments making the compound proposition false. For $n=2$ this is $2^2-1=3$, exactly as you discovered from your truth tables.
