Consider the problem of constructing a latin square of order $N$, using only row and column values from a given word list ($W$) containing some subset of the $N!$ possible word values.
For example, for $N = 4$, given word list {1234, 2341, 3412, 4123}, we can obtain 4 latin squares:
$$\begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3
\end{bmatrix}
\begin{bmatrix}
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4
\end{bmatrix}
\begin{bmatrix}
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1
\end{bmatrix}
\begin{bmatrix}
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2
\end{bmatrix}$$
The question naturally arises - for given $N$, how large must $|W|$ be in order to guarantee that at least one latin square is obtainable?
I'll call this minimum word list size $M_N$.
So, for given $N$, $M_N$ is the least integer such that a Latin square is obtainable from all $W$ with $|W|=M_N$.
We have $M_2 = 2, M_3 = 5$, and $M_4 = 18$, these are easily established. As a proportion of the $N!$ words applicable, these are 1, 5/6 and 3/4 respectively.
For $N=5$, I have found a set of 96 words which doesn't yield a latin square, so $M_5 > 96$.
Note that $18 = 4! - 3!$, $96 = 5! - 4!,$ and that we can always construct a set $W_N$ that contains all $N!$ words except those beginning with $1$ (or any other value) from which no Latin squares can be obtained.
So $M_N = N! - (N-1)! + k_N$, with $k_N > 0$, and we suspect, but have yet to prove, that $k_N = 1$ for all $N$.