The compound hepta-2,5-dien-4-ol, CH3CH=CH-CH(OH)-CH=CHCH3, can exist in several stereoisomeric forms because of its two double bonds and its…

Each double bond can be $E$ or $Z$, giving combinations $(E,E)$, $(Z,Z)$ and $(E,Z)$ (the last equals $(Z,E)$ by symmetry of the constitution). In $(E,E)$ and $(Z,Z)$ the two groups on $\mathrm{C4}$ are identical, so $\mathrm{C4}$ is not a stereocentre and each is a single achiral compound with a mirror plane through $\mathrm{C4}$: $2$ stereoisomers, both optically inactive. In $(E,Z)$ the two alkenyl groups on $\mathrm{C4}$ are different ($E$-propenyl versus $Z$-propenyl), so $\mathrm{C4}$ carries four different groups ($\mathrm{H}$, $\mathrm{OH}$, and two diastereomeric alkenyl groups) and is a stereocentre; this gives a pair of enantiomers, $2$ stereoisomers, both optically active. Thus $N=2+2=4$, $M=2$, and $N+M=6$.