Methylcyclopentane undergoes free-radical monochlorination. The total number of distinct monochloro products obtained, counting every stereoisomer…

Methylcyclopentane has four types of hydrogen: on the methyl group, on C1 (the tertiary ring carbon), on C2/C5 and on C3/C4. Chlorination at $\mathrm{CH_3}$ gives (chloromethyl)cyclopentane, with no stereocentre: 1 product. Chlorination at C1 gives 1-chloro-1-methylcyclopentane, which is achiral: 1 product. Chlorination at C2 gives 1-chloro-2-methylcyclopentane with two stereocentres and no symmetry: cis and trans, each chiral, so $2^2=4$ stereoisomers. Chlorination at C3 gives 1-chloro-3-methylcyclopentane; since the two substituents differ there is no mirror plane in either the cis or trans isomer, giving 4 stereoisomers again. Total $=1+1+4+4=10$.