In the arithmetic sense,
out = (in ≥ 1 ∧ in ≤ 10)and
out = ¬(in < 1 ∨ in > 10)are exactly the same. Here, I'm using
∧ for OR,
∨ for AND, and
¬ for NOT, as is common.
Is one of the above more correct than the other? No, both of them are equally exactly correct.
But, if you can ONLY use AND (
∨) and OR (
∧), and cannot use NOT (
¬), then you'd be limited to the first one, using OR.
I personally don't like it when I'm given such artificial limitations; there should always be a stated reason why, because often there is an even better
overall solution that makes the artificial limitation completely irrelevant.
Let's examine the truth table by splitting the input into four binary inputs:
$$\begin{array}{cc|c}
\text{Input} & D C B A & \text{Output} \\
\hline
0 & 0 0 0 0 & 0 \\
1 & 0 0 0 1 & 1 \\
2 & 0 0 1 0 & 1 \\
3 & 0 0 1 1 & 1 \\
4 & 0 1 0 0 & 1 \\
5 & 0 1 0 1 & 1 \\
6 & 0 1 1 0 & 1 \\
7 & 0 1 1 1 & 1 \\
8 & 1 0 0 0 & 1 \\
9 & 1 0 0 1 & 1 \\
10 & 1 0 1 0 & 1 \\
11 & 1 0 1 1 & 0 \\
12 & 1 1 0 0 & 0 \\
13 & 1 1 0 1 & 0 \\
14 & 1 1 1 0 & 0 \\
15 & 1 1 1 1 & 0 \\
\end{array}$$
If there are more significant binary digit inputs than the above four, any one of them being one means the output must be zero.
The corresponding
Karnaugh map would be 4×4. For example,
$$\begin{array}{c|cccc}
~ \diagdown & BA & ~ & ~ & ~ \\
DC & 01 & 11 & 10 & 00 \\
\hline
00 & 1 & 1 & 1 & 0 \\
01 & 1 & 1 & 1 & 1 \\
11 & 0 & 0 & 0 & 0 \\
10 & 1 & 0 & 1 & 1 \\
\end{array}$$
One way to write this would be
out = ¬( (D ∧ C) ∨ (D ∧ B ∧ A) ∨ (¬(D ∨ C ∨ B ∨ A))which uses both AND (
∧) and OR (
∨) operations, as well as NOT (
¬). The
(D ∧ C) refers to the entire third row, the
(D ∧ B ∧ A) refers to the second column in the third and fourth rows, and the
¬(D ∨ C ∨ B ∨ A) refers to the fourth column on the first row. These three cover all zeros in the K-map, so OR'ing them together and inverting the result, yields the output.
(There are different methods for solving K-maps, and perhaps this one isn't the one you arrive at using what you have in your proverbial "toolbox". That's okay; I pulled this one way out of my proverbial backside, using my experience and observations in minimizing binary expressions in programming as my "toolkit".)
So, as you can see, the underlying question,
"how to decide what to use" is more of an optimization problem: you decide what to use, based on what gives you the expression you prefer. How you compare different valid expressions, is up to you.