The number of isomers of $C_3{H}_{5}B{r}_3$  is (including stereoisomers)

Final answer: There are 4 distinct isomers (including stereoisomers) of C₃H₅Br₃.
Why only 4? 

Start from propane, CH₃–CH₂–CH₃. We must replace three H atoms by three Br atoms. Look at how the three Br atoms can be distributed over the three carbon atoms, remembering the two end CH₃ groups are equivalent (mirror symmetry of propane).

Only these distinct substitution patterns (connectivities) are possible:

All three Br on the same terminal carbon → 1,1,1-tribromopropane
Structure: C(Br)₃–CH₂–CH₃
(All three Br on one end carbon.)

Two Br on one terminal carbon and one Br on the middle carbon → 1,1,2-tribromopropane
Structure: C(Br)₂–CH(Br)–CH₃
(Two on an end, one on the middle carbon.)

Two Br on the middle carbon and one Br on a terminal carbon → 1,2,2-tribromopropane
Structure: CH₃–C(Br)₂–CH₂Br (locants arranged to give lowest numbers: 1,2,2)
(Both hydrogens of middle carbon replaced + one on an end.)

One Br on each carbon → 1,2,3-tribromopropane
Structure: Br–CH₂–CH(Br)–CH₂–Br (one Br on each carbon)

These four patterns are the only connectivity-wise different arrangements once you account for symmetry (the two ends of propane are equivalent, so many naive placements reduce to the same isomer).

About stereoisomers

None of the four constitutional isomers above gives a separate pair of non-superposable stereoisomers (enantiomers) in this particular set (no stable chiral center with four different groups is produced here with three identical substituents Br). So the count does not increase beyond 4.

(You can check each structure: there is no carbon attached to four different groups which would create an enantiomeric pair.)

Therefore

Total distinct isomers (including any stereoisomers) = 4.