Newtonian approximation in cosmology

In Peeble's "Large scale structure of the Universe" the Newtonian approximation in cosmology is examined by looking at a transformation that makes the metric locally Minkowski. On pg 38 it is stated that

To make $g_{ij,kl}=0$, we would add 10$\times$10 equations for a total of 150 to be satisfied by the choice of 56+80=136 transformations.

The 56 is explained a couple of lines earlier but I am having trouble seeing where the 80 comes from. I would very much appreciate any help with how to derive that an additional 80 transformations are needed in the above quote.

During coordinate transformation $x^i(y^j)$ the change of components of metric is expressed through first derivatives $\frac{\partial x^i}{\partial y^j}$, first derivatives of metric $g_{ij,k}$ will contain the second derivatives of $x^i$, while expression for the second metric derivatives will contain the third derivatives of $x^i$.
So, 80 is the number of third derivatives $$\frac{\partial ^3 x^i}{\partial y^j \partial y^k \partial y^l}$$ involved in varying the second derivatives of metric $g_{ij,kl}$ during coordinate transformations $x^i(y^j)$.
It is composed as $80 = 4 \times 20$, where 4 is the number of values for index $i$ in numerator, while 20 is the number of unordered triples $\{j,k,l\}$ of indices in denominator. In turn $20=4+4\times 3 + 4$, first term is the number of triples with all indices identical, second -- with two identical indices and third with all three indices distinct.
So, by choosing coordinate transformation $x^i(y^j)$ we could make the metric locally Minkowskian ($g_{ij}(x^i_0)=\eta_{ij}$) and also turn the first derivatives of metric to zero ($g_{ij,k}=0$). However we cannot generally also turn second derivatives to zero.