We determine the number of geometrical isomers for each complex one by one.

1. $[PtCl_2(NH_3)_2]$

Step 1: Identify geometry

Platinum in such complexes is typically square planar.

So the complex is of type $[MA_2B_2]$ in a square planar arrangement.

Step 2: Geometrical isomers possible

For square planar $[MA_2B_2]$, two geometrical isomers are possible:

1. cis: like ligands adjacent
2. trans: like ligands opposite

Hence, number of geometrical isomers = 2.

2. $[Ni(CO)_4]$

Step 1: Identify geometry

$[Ni(CO)_4]$ is tetrahedral.

Step 2: Check geometrical isomerism

In a tetrahedral complex with all four ligands identical, there is only one arrangement.

Hence, number of geometrical isomers = 0.

3. $[Ru(H_2O)_3Cl_3]$

Step 1: Identify geometry

This is a six-coordinate complex, so it is octahedral.

It is of type $[MA_3B_3]$.

Step 2: Geometrical isomers possible

For octahedral $[MA_3B_3]$, two geometrical isomers are possible:

1. fac (three identical ligands on one face)
2. mer (three identical ligands in a meridional plane)

Hence, number of geometrical isomers = 2.

4. $[CoCl_2(NH_3)_4]^+$

Step 1: Identify geometry

This is a six-coordinate complex, so it is octahedral.

It is of type $[MA_4B_2]$.

Step 2: Geometrical isomers possible

For octahedral $[MA_4B_2]$, two geometrical isomers are possible:

1. cis
2. trans

Hence, number of geometrical isomers = 2.

Final sequence

Thus the numbers are:

$2,,2,2$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.