Let, we have two rows. First row has $n$ particles $A,B,C,D$ and other row has $A',B',C',D'$ particles, which are antiparticles of the previous ones. Also, they have colours$R,B,G,V$ whereas antiparticles have anti colours. Then, number of colour less arrangements are $$T=\sum_{a+b+c+d=n} \binom{n}{a,b,c,d}^2$$
The same arrangement can be thought as $2$-groups $A,B$ and $C,D$. We choose $k$ places on both rows for $A,B$ group, in $\binom{n}{k}^2$ ways. So, this can be made colour less in $$\sum_{x+y=k} \binom{n}{x,y}^2=\binom{2k}{k}$$ ways. Similarly the $C,D$ group can be made colour less in $\binom{2n-2k}{n-k}$ ways.
Hence, total $$T=\sum_{k=0}^{n} \binom{n}{k}^2\binom{2k}{k}\binom{2n-2k}{n-k}$$ ways. To be honest, I have given a combinatorial shape to the way the equality can be obtained.