Butan-2-one, $\mathrm{CH}{}_3\mathrm{COCH}{}_2\mathrm{CH}{}_3$, can enolise toward either alpha carbon. The total number of enol tautomers, counting geometrical (cis/trans) isomers as separate, is

Enolising toward the methyl group gives $\mathrm{CH}{}_2=\mathrm{C}(\mathrm{OH})\mathrm{CH}{}_2\mathrm{CH}{}_3$ (but-1-en-2-ol), with no geometrical isomerism. Enolising toward the methylene gives $\mathrm{CH}{}_3\mathrm{C}(\mathrm{OH})=\mathrm{CHCH}{}_3$ (but-2-en-2-ol), whose double bond carries two different groups on each carbon, so E and Z forms exist. Total $=1+2=3$.