1. Identify the complex ion

The compound is $[Pt(NH_3)_4Cl_2]Br_2$ So the coordination entity is $[Pt(NH_3)_4Cl_2]^{2+}$ and $Br^-$ ions are outside the coordination sphere.

2. Oxidation state and geometry of Pt

Let oxidation state of Pt be $x$. Since $NH_3$ is neutral and $Cl^-$ contributes $-2$ inside the bracket, $x-2=+2$ $x=+4$ Thus metal is $Pt^{4+}$.

For coordination number $6$, platinum(IV) complexes are generally octahedral. So we need isomers of an octahedral complex of type $[MA_4B_2]$ with $A=NH_3, \quad B=Cl^-$

3. Geometrical isomerism of $[MA_4B_2]$

An octahedral complex of type $[MA_4B_2]$ shows 2 geometrical isomers:

• cis
• trans

So within the coordination sphere, there are 2 isomers.

4. Check for optical isomerism

For octahedral $[MA_4B_2]$, neither cis nor trans form is optically active. So optical isomers = 0 extra.

Hence, stereoisomers of the coordination entity = 2.

5. Ionisation isomerism?

Ionisation isomerism occurs when a ligand inside the coordination sphere can exchange with the counter ion outside.

Here, inside we have $Cl^-$ and outside we have $Br^-$. Therefore possible ionisation isomers are:

• $[Pt(NH_3)_4Cl_2]Br_2$
• $[Pt(NH_3)_4ClBr]ClBr$
• $[Pt(NH_3)_4Br_2]Cl_2$

These are three different ionisation forms.

6. Count geometrical isomers for each ionisation form

(i) $[Pt(NH_3)_4Cl_2]Br_2$

Coordination sphere type: $[MA_4B_2]$

Geometrical isomers = $2$ (cis, trans)

(ii) $[Pt(NH_3)_4ClBr]ClBr$

Coordination sphere type: $[MA_4BC]$

For octahedral $[MA_4BC]$, there are also $2$ geometrical isomers:

• $B$ and $C$ cis
• $B$ and $C$ trans

So geometrical isomers = $2$

(iii) $[Pt(NH_3)_4Br_2]Cl_2$

Coordination sphere type: $[MA_4B_2]$

Again geometrical isomers = $2$ (cis, trans)

7. Total isomers

Therefore total possible isomers $=2+2+2=6$

8. Final answer

The total number of possible isomers is $\boxed{6}$

This matches the stored correct answer.