boolean equation to truth table My question is to make a truth table from a given boolean equation.
The boolean equation is $M =  A'B’C +  A’BC’  +  ABC’ +  ABC$
I get the following results but I think I did something wrong. Any help or pointers would be appreciated.
The output of the truth table was: $00110101$
 A: You indeed did something wrong.
If you look at the term $M=A′B′C+A′BC′+ABC′+ABC$, then $M$ is $1$ iff any of the terms is $1$. Each term is true iff every of its factor is $1$. And $A'$ is $1$ if $A$ is $0$.
Let's look at a smaller expression (which also saves me typing):
$X = A'B' + A'B + AB$
Let's first consider the first term, $A'B'$. This is $1$ if both $A'$ and $B'$ are $1$. But $A'=1$ if $A=0$, and $B'=1$ if $B=0$. So the only case when $A'B'$ is $1$ is when $A=B=0$.
Now let's look at the second term, $A'B$. Again, this is $1$ iff $A'=B=1$. But $A'=1$ iff $A=0$. Therefore the only way $A'B$ is $1$ is iff $A=0$ and $B=1$.
You might see a pattern now: In each term, all the primed factors have to be $0$ and all non-primed $1$. So let's verify it at the third term: $AB$. Both are unprimed, so both have to be $1$. And indeed, that's the case.
Now $X$ becomes $1$ if any of the terms is $1$.  Therefore for $X$ we get the truth table:
$$\begin{array}{cc|c|c}
A & B & \text{matching term} & X\\
\hline
0 & 0 & A'B' & 1\\
0 & 1 & A'B & 1\\
1 & 0 & \text{---} & 0\\
1 & 1 & AB & 1
\end{array}$$
As you see, $X=1$ if there's a matching term in $X$, and $X=0$ otherwise.
Remark: This can also done in reverse. Because your truth table has an entry $1$ in the rows $010$, $011$, $101$ and $111$, your truth table would be correct for the expression $Y=A'BC'+A'BC+AB'C+ABC$.
With this information, you should now be able to get the correct truth table also for $M$.
