The total number of stereoisomers (including meso forms) of pentane-2,3,4-triol, CH3CH(OH)CH(OH)CH(OH)CH3, is

Three carbons, C-2, C-3, C-4, bear $\mathrm{OH}$. C-2 and C-4 are constitutionally equivalent, so $2^3 = 8$ overcounts. When C-2 and C-4 are $(R,R)$ or $(S,S)$, C-3 has two identical-descriptor arms and is not a stereocentre: this gives the enantiomeric pair $(2R,4R)$/$(2S,4S)$ = 2 isomers. When C-2 and C-4 are $(R,S)$, C-3 is pseudoasymmetric, giving two achiral meso forms ($3r$ and $3s$). Total $=2+2=4$.