The total number of stereoisomers (counting each enantiomer separately, and counting constitutional isomers as different compounds) of all aldehydes…

Aldehydes are $\mathrm{R{-}CHO}$ with $\mathrm{R}=\mathrm{C_5H_{11}}$. There are $8$ pentyl groups: n-pentyl, 2-pentyl, 3-pentyl, 2-methylbutyl, 3-methylbutyl, 1,1-dimethylpropyl, 1,2-dimethylpropyl and neopentyl. Of the corresponding aldehydes, three have a stereocentre: 2-methylpentanal, 3-methylpentanal and 2,3-dimethylbutanal. So aldehydes give $8+3=11$ stereoisomers. Ketones $\mathrm{R{-}CO{-}R'}$ with $\mathrm{R}+\mathrm{R'}$ together $\mathrm{C_5}$: methyl with the four butyl groups (4 ketones) and ethyl with propyl or isopropyl (2 ketones), a total of $6$. Only 3-methylpentan-2-one is chiral, so ketones give $6+1=7$. Total $=11+7=18$. Answer: 18.