We do case work and, if applicable, within each case further divided two sub-cases depending on whether the two $o$s present. If both $o$s present, it will introduce a duplicate factor of $2$.

• $5$-letter word: $\frac{P_5^5}{2!}=60$
• $4$-letter word: $P_4^4 + \frac{C_3^2\times P_4^4}{2!}=60$
• $3$-letter word: $P_4^3 + \frac{C_3^1\times P_3^3}{2!}=33$
• $2$-letter word: $P_4^2 + 1=13$
• $1$-letter word: $4$

Therefore totally $\boxed{170}$ words can be constructed.