Count the distinct enol forms of the compound shown, counting E and Z geometrical isomers separately. How many are there?

The compound shown is pentan-3-one. Both alpha carbons are equivalent CH${}_2$ groups, so there is only one constitutional enol, $\mathrm{CH}{}_3\mathrm{CH}=\mathrm{C}(\mathrm{OH})\mathrm{CH}{}_2\mathrm{CH}{}_3$. Its C=C carries different groups at each end, so it exists as E and Z isomers, giving 2 enol forms in total.