How many stereoisomers (enantiomers plus any meso form) are possible for 2,3-dibromobutane, $\mathrm{CH}{}_3-\mathrm{CHBr}-\mathrm{CHBr}-\mathrm{CH}{}_3$?

The molecule has two stereocentres, so the $2^n$ rule gives 4 combinations. Because the two ends of the chain are identical, the (2R,3S) and (2S,3R) combinations are the same achiral meso compound. The stereoisomers are (2R,3R), (2S,3S) and meso, so the number is 3.