Transversals of Latin Squares According to this thesis, page $28$, the following Latin Square has $3$ $0$-s transversals: $$\begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 2 & 4 & 1 & 5 & 3\\ 3 & 5 & 4 & 2 & 1\\ 4 & 1 & 5 & 3 & 2\\5 & 3 & 2 & 1 & 4\end{bmatrix} \implies \begin{bmatrix}1 & & & &\\ & & & 5 & \\ & & 4 & & \\ & & & & 2\\ & 3\end{bmatrix}, \begin{bmatrix}& & & 4 &\\2 & & & &\\ & & & & 1 \\ & & 5 & &\\ & 3\end{bmatrix}, \begin{bmatrix}& & & &5\\& & 1 & &\\ & & & 2 &\\4 & & & &\\& 3\end{bmatrix}.$$ The definition of a $0$-s transversal for a Latin square of order $n$ is 

a set of $n$ ordered triples such that the first and second entries are the rows and columns respectively in which the values $1,\ldots,n$ occur exactly once and the third entry of the triple is the value, of which there are $n$ distinct values.

Basically, we need to visit each row and column only once and we must have $5$ distinct symbols at the end. I can represent each transversal as $$t_1 = \{(1,1,1),(2,4,5),(3,3,4),(4,5,2),(5,2,3)\}$$ $$t_2 = \{(1,4,4),(2,1,2),(3,5,1),(4,3,5),(5,2,3)\}$$ $$t_3 = \{(1,5,5),(2,3,1),(3,4,2),(4,1,4),(5,2,3)\}$$ So why are these the only three? How do I know there are only three of them?

A latin square is represented as $t_i=(r,c,s)$ which means that r represents row 
of the latin square, c represents column of the latin square and s is the number
whose location is $row=r\;\;and\;\;column=c$.
