The compound CH3CH=CH-CH(OH)-CH=CH-CH3 (hepta-2,5-dien-4-ol) is considered with all its possible configurational stereoisomers, counting enantiomers…

There are two $\mathrm{C{=}C}$ bonds, each of which can be $E$ or $Z$, and one carbinol carbon, $\mathrm{C_4}$, carrying $\mathrm{H}$, $\mathrm{OH}$ and two propenyl arms. Take the combinations of double-bond geometry. In the $(E,E)$ and $(Z,Z)$ isomers the two arms at $\mathrm{C_4}$ are identical, so $\mathrm{C_4}$ is not stereogenic and each molecule is achiral with a mirror plane. This gives 1 isomer each, so 2 in total. In the $(E,Z)$ combination one arm is an $(E)$-propenyl and the other a $(Z)$-propenyl group. These arms are diastereomorphic (not mirror images), so $\mathrm{C_4}$ carries four different groups and is a genuine stereocentre; by the CIP rule $Z>E$ it can be $R$ or $S$. The $(E,Z)$ molecule has no mirror plane, so it exists as a pair of enantiomers, $(4R)$ and $(4S)$. The $(Z,E)$ combination is the same molecule as $(E,Z)$ read from the other end, so it adds nothing new. Total $=1+1+2=4$.