The number of stereoisomers (counting enantiomers separately) of the compound CH3CH=CH-CH(OH)-CH=CH-CH3 (hepta-2,5-dien-4-ol) is

There are two stereogenic double bonds (C2=C3 and C5=C6) and a carbinol carbon C4. Treat the double-bond geometries first. For $(E,E)$ and $(Z,Z)$ the two groups on C4 are identical in constitution and configuration, so C4 is not a stereocentre and each gives one achiral (meso-like) compound. For $(E,Z)$ the two alkenyl arms differ in configuration, so C4 carries four different groups and is a true stereocentre; the $(E,Z)$ and $(Z,E)$ descriptions are the same compound with the numbering reversed, so we get only one constitution-geometry pair but two enantiomers, $(4R)$ and $(4S)$. Total $=1+1+2=4$. The naive $2^3=8$ overcounts, and 3 forgets the enantiomeric pair.