Number of diastereomer of the given compound is:

Number of chiral C-atoms = 3, (n = 3, odd number and terminal groups are same).
Number of total isomers :$2^{\mathrm{n}-1}=2^{3-1}=4$.
Number of optically inactive (meso forms) $=2^{\frac{n-1}{2}}=1$
Number of optically active forms = 4 - 1 = 3
Hence number of diastereomers = 3.

II and III are same. Check after rotating to 180°
.'. Diastereomer pairs are
1. I and II or III 2. I and IV
3. II or III and IV