Combinatorics problem

Using condition (i), $R_{abcc}=-R_{abcc} \implies R_{abcc}=0$ So, for independent elements, the last two indices should be different. Again, from condition (i), we have, for example, $R_{1234}=-R_{1243}$ If we know the value of $R_{1234}$ , we know the value of $R_{1243}$ straight away. Only one of them is independent.

Considering condition (i) only, we can conclude, for every combination (obviously distinct numbers) of the last two indices, we have one independent element.

And there are $\frac{N(N-1)}{2!}$ number of such combinations.

Now considering condition (ii), $R_{aacd}=0$ , so $R_{aacd}$ is not independent. Again, for every combination (obviously distinct integers) of the first two indices, we have one independent element.

And there are $\frac{N(N-1)}{2!}$ number of such combinations.

So, considering condition (i) and (ii), there are $\frac{N(N-1)}{2!}\cdot\frac{N(N-1)}{2!}$ independent elements.

Now, let's consider condition (iii). According to condition (iii), if we interchange the first couple of indices with the last couple of indices, the element remains the same.

But if $(a,b)=(c,d)$ , for example, $R_{1212}=R_{1212}$ , this is an identity.

There are ${}^NC_2=\frac{N(N-1)}{2!}$ such identities.

Otherwise, for each independent element, there is a corresponding dependent element. So considering condition (i), (ii) and (iii), there are

$\frac{1}{2}\left(\frac{N(N-1)}{2!}\cdot\frac{N(N-1)}{2!}-\frac{N(N-1)}{2!}\right)+\frac{N(N-1)}{2!}=\frac{1}{2}\cdot\frac{N(N-1)}{2!}\left(\frac{N(N-1)}{2!}+1\right)$

independent elements.

Now, consider condition (iv). We have to count the number of independent equations that condition (iv) provides.

If $a=b$ , we have,

$R_{aacd}+R_{adac}+R_{acda}=0$

$\implies R_{adac}=-R_{acda}$ ... (v) [ $R_{aacd}=0$ according to condition (ii)]

Equation (v) can also be derived using condition (i) and (iii). Thus, when $a=b$ , condition (iv) does not provide any new independent equation. Same can be shown for $a=c$ , $a=d$ , $b=c$ , $b=d$ and $c=d$ .

Again,

$R_{abcd}+R_{adbc}+R_{acdb}=0$

$\implies-R_{bacd}+R_{bcad}+R_{dbac}=0$ [Using condition (ii) and (iii)]

$\implies-R_{bacd}-R_{bcda}-R_{bdac}=0$ [Using condition (i) and (ii)]

$\implies R_{bacd}+R_{bcda}+R_{bdac}=0$ ... (vi)

So if equation (v) is satisfied (also condition (i), (ii) and (iii) ), equation (vi) is automatically satisfied.

In the same way, it can be shown that the following equations are also satisfied:

$R_{cabd}+R_{cdab}+R_{cbda}=0$

& $R_{dabc}+R_{dcab}+R_{dbca}=0$

So, for $(a,b,c,d) = (1,2,3,4) , (1, 2, 4, 3) , (1, 3, 2, 4) ...$ any permutation of $(1,2,3,4)$ , we have only one independent equation from condition (iv).

Thus, for each combination of four distinct integers, we have only one independent equation. So we get ${}^NC_4$ independent equations.

So considering all the conditions, we have $\frac{1}{2}\cdot\frac{N(N-1)}{2!}\left(\frac{N(N-1)}{2!}+1\right)-{}^NC_4=\frac{N^2(N^2-1)}{12}$ independent elements.