The Boolean expression AB + AC̅ + BC simplifies to

The question asks us to simplify the given Boolean expression: $AB+AC\overline{B}+BC$

Simplifying the Boolean Expression

We can simplify this expression using the rules of Boolean algebra. Let's focus on simplifying the terms involving $AC\overline{B}$ and $BC$.

Consider the part $AC\overline{B}+BC$. We can factor out $C$: $AC\overline{B}+BC=C(A\overline{B}+B)$

Now, let's simplify the expression inside the parenthesis: $A\overline{B}+B$. We can use the Boolean algebra identity $X+\overline{X}Y=X+Y$. Let $X=B$ and $Y=A$. Applying this identity, we get: $B+\overline{B}A=B+A$

Substituting this back into our expression: $C(A\overline{B}+B)=C(B+A)$

Using the distributive law ($C(B+A)=CB+CA$), we get: $CB+CA$ Which is equivalent to: $BC+AC$

So, we have shown that $AC\overline{B}+BC$ simplifies to $AC+BC$.

Now, substitute this simplified part back into the original expression: $AB+AC\overline{B}+BC=AB+(AC\overline{B}+BC)=AB+AC+BC$

The expression $AB+AC+BC$ can be further simplified. Let's consider the original expression again: $AB+AC\overline{B}+BC$. The simplification derived ($AC+BC$) matches the first option. While further simplification of $AB+AC+BC$ might be possible, the structure $AC\overline{B}+BC$ directly leads to one of the options.

Therefore, the Boolean expression $AB+AC\overline{B}+BC$ simplifies to $BC+AC\overline{B}$.