The total number of stereoisomers (counting each enantiomer separately) possible for the compound CH3CH=CH-CH(OH)-CH=CHCH3 is

There are two stereogenic double bonds and one carbinol carbon. The carbinol carbon bears H, OH and the two $\mathrm{CH{=}CHCH_3}$ arms. It is a stereocentre only when the two arms differ, i.e. when the double-bond geometries are $E,Z$. Cases: (E,E) has identical arms, so no stereocentre and the molecule is achiral: 1 isomer. (Z,Z): likewise 1 isomer. (E,Z): the carbinol carbon is a true stereocentre, giving two enantiomers (R and S): 2 isomers. A combinatorial enumeration with the symmetry equivalence $(e_1,e_2,s)\sim(e_2,e_1,-s)$ confirms 4 distinct stereoisomers in total. Answer: 4.