Why does the lowered Riemann tensor only have 20 independent components for the Schwarzschild metric? I have seen quite a bit online about how there are only 20 independent components for the (lowered) Riemann tensor $R_{abcd}$ for the Schwarzschild metric. I've been told this follows from the symmetries of the tensor, i.e.:
$R_{abcd}=-R_{bacd}=-R_{abdc}=R_{badc}$ and $R_{abcd}=R_{cdab}$
Now if the indices in the tensor can all run from 1 to 4, then $R_{abcd}$ has 256 components. These symmetries seem to reduce our need to calculate components, but why only 20?
Note: I am aware there are some similar questions on the stack exchange. I have read these, but none of them very clearly explained this specific point, so I decided to ask it directly so hopefully I can get my head around it.
 A: The comment by FrodCube points to the December 1997 Lecture Notes on General Relativity by Sean M. Carroll equation (3.85) which when evaluated for n=4 gives 20. 
For those who don't want to open the link/it does not exist/want to see it quickly, my answer is for them. 
First note that $^nC_{k}$ is given by (https://en.wikipedia.org/wiki/Binomial_coefficient), 
\begin{equation}
^nC_{k} = \frac{n!}{k! (n-k)!}
\end{equation} 
The number of independent components of Riemann tensor ($IRC$) in $d$ space-time dimensions is simply, 
\begin{equation}
 IRC = \frac{~^dC_{2} (^dC_{2} + 1)}{2} - ~^dC_{4}
\end{equation} 
A: Just to summarise the calculation FrodCube mentioned, $R_{abcd}\ne 0$ only when $a\ne b$ and $c\ne d$. In $n$-dimensional spacetime there are $\binom{\binom{n}{2}+1}{2}$ ways to choose the pairs $ab,\,cd$ (they're allowed to be the same, hence the $+1$). We then lose $\binom{n}{4}$ degrees of freedom to the first Bianchi identity $R_{abcd}+R_{acdb}+R_{adbc}=0$. A bit of algebra says that leaves $\frac{1}{6}\binom{n^2}{2}$ DOFs, which for $n=4$ is $20$ as required.
