The pairs $(f(2),f(3))$ with $f(2)+f(3)=5$, $f(3)\le 4$, distinct positive integers in $\{1,\dots ,6\}$ are: $(4,1)$, $(1,4)$, $(3,2)$, $(2,3)$. Four valid cases.

In each case, $f(2)$ and $f(3)$ are fixed. The constraint $f(1)\ge 3$ leaves three values among $\{3,4,5,6\}$ that are still available (since two have been used by $f(2)$ and $f(3)$, and in each case exactly one of the remaining values is below $3$). Verification: e.g. for $(f(2),f(3))=(4,1)$, the unused set is $\{2,3,5,6\}$, and those $\ge 3$ are $\{3,5,6\}$, giving $3$ choices.

After choosing $f(1)$, three values remain for $f(4),f(5),f(6)$: $3!=6$ permutations.

Per case: $3\times 6=18$. Total $=4\times 18=72$.