Tetrahydroxyhexanedioic acid (saccharic acid type), $\mathrm{HOOC}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{COOH}$, has four stereocentres and a symmetric constitution. How many of its stereoisomers are optically active? (Count each enantiomer separately.)

For a symmetric chain with an even number $n=4$ of stereocentres the total number of stereoisomers is $2^{n-1}+2^{n/2-1}=8+2=10$. The meso forms number $2^{n/2-1}=2$. The optically active isomers are therefore $10-2=8$ (four enantiomeric pairs). This count was confirmed by enumerating all $2^4=16$ configurations and merging those related by the symmetry of the molecule.