Independent terms

What we see from the given specialties that we can get $10$ independent elements for each $(i,j); (k,l); (m,n)$ from first three properties. Now, say $T_{ijklmnpq}=T_{{\alpha}{\beta}{\gamma}{\delta}}$ . Now, $\alpha, \beta, \gamma$ all can have $\frac{(4*4-4)}{2}+4=10$ independent points and $\delta$ can have $4*4=16$ . Now, from $4,5$ we can see whenever at least two of $\alpha, \beta, \gamma$ meaning, at least any two of $(i,j); (k,l); (m,n)$ are equal $T_{ijklmnpq}=0$ . Hence, we can chose independent points for { $ijklmn$ } in $\binom{10}{3}=120$ ways and in additional to this we can choose $16$ point for $pq$ . hence, total $120*16=1920$ ways. So, we have total 1920 ways to choose independent components( considering each connection of 4 points of $\alpha$ , $\beta$ , $\gamma$ , $\delta$ an element) for this $(0,8)$ tensor $T$ . hence, our desired answer is $1920+100=2020$ .