Counting members of equivalence classes of injective functions 
Let $A=\{f:\{1,2,3,4\} \to \{1,2,3,4,5,6,7,8\}, f$ injective$\}$. Let $R$ be a relation on $A$ such that $f_0Rf_1 \iff f_0(1)+f_0(2) = f_1(1)+f_1(2)$. If $h \in A, h(n)=n+2$, what's the cardinality of its equivalence class?

My approach to this exercise was the following. We know $h(1) + h(2) = 7$. We also know that there are ${6}\choose{5}$ ways to distribute seven indistinguishable balls into two boxes such that both have at least one ball. But this is equivalent to counting the injective mappings $\{1,2\} \to \{1,2,3,4,5,6\}$ such that the elements of the image add up to 7.
Hence the number of injections appropriately related to $h$ is ${{6}\choose{5}} \cdot 6 \cdot 5$.
Is this correct?

Let $f$ in the equivalence class of $h$, i.e. $f$ is injective and $f(1)+f(2)=7$. First, you have 6 options for $f(1)$, then $f(2):=7-f(1)$ is determined. So far, we have taken two values$(f(1),f(2))$ from $\{1,\dots, 8\}$, so we are left with $6$ possible choices for $f(3)$ and finally, $5$ possible choices for $f(4)$. So, in total is $6*6*5$. 
