These are two problems of the Velleman's How to prove book in which it has been asked to find if $R$ is an partial order or/and an total order:
$R1 = \{(x,y) \in A \times A \mid \text{the word $y$ occurs at least as late in alphabetical order as the word $x$} \}$
$R2 = \{(x,y) \in A \times A \mid \text{the first letter of the word $y$ occurs at least as late in the alphabet as the first letter of the word $x$}\}$
where $A$ is the set of all words of English.
Now I conclude that $R1$ is not a total order because $(haskell, go) \notin R$. Also for the set $R2$ it is not a partial order because of an element like $(aba, aa)$. But the solution presented here claims otherwise. So am I missing something here ?
A total order requires that either $(x,y)$ or $(y,x)$ be part of the relation. It is similar to $\le$ on the naturals. We don't have $2 \le 1$, but we do have $1 \le 2$, so it could be (and is) a total order. Similarly, your example (haskell,go) is not part of the order, but (go,haskell) is. For the second, a partial order has to be transitive and antisymmetric. In this case, both (aa,aba) and (aba,aa) are part of the relation. Since aa $\neq$ aba, it is not antisymmetric and this is not a partial order.