The ketone shown has two different alpha carbons. Counting only enols with the OH on the former carbonyl carbon and counting E and Z isomers separately, how many distinct enols can it give?

The ketone is 2-methylpentan-3-one. Enolisation toward the CH2 gives CH3CH=C(OH)CH(CH3)2 as E and Z. Enolisation toward the CH gives (CH3)2C=C(OH)CH2CH3, with two identical methyl groups on one alkene carbon, so no E/Z. Total = 2 + 1 = 3.