# Counting the number of Einstein field equations

This might be a very naive question but I do not understand how the Einstein's Field Equations give us enough equations to solve for the metric in certain problems. In general, the EFEs have 10 independent equations because both the Einstein tensor and the Stress-Energy are symmetric. This is an adequate number of equations to determine the components of the 10 components of the metric if the stress energy tensor is either a function of the metric or is already known like in the case of the Schwarzschild solution. But as far as I know, we do not know the components of the stress-energy tensor beforehand most of the time. For instance, consider the stress-energy tensor for a collapsing dust: $$T^{\mu \nu} = \rho U^{\mu}U^{\nu}$$ Here, the velocity field of the dust is not known at all points in time and needs to be solved for (with an initial velocity field perhaps). But now we have 4 unknowns from the four velocity and 10 unknowns from the metric. However, we only have 10 Einstein field equations.

Another statement that confuses me is that the condition that the Einstein tensor and the stress energy be divergenceless (which gives us four equations) is inbuilt in the tensors and therefore reduces the number of independent equations from 10 to 6. It seems to me that the number of equations are still 10; if the divergenceless condition gives us no additional information, it should contribute no additional equations but not reduce the number of independent equations in the EFEs.

I understand that the remaining 6 equations are adequate to solve for the metric uniquely (up to a change in coordinates), but I just don't understand how we got here. If there are only 6 independent equations, how do we even solve arbitrary dust problems where there is no symmetry (let's say I were to program this into a computer) and there are 14 unknowns but only 6 equations?