The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1,2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is

$3 \mathrm{digit} \mathrm{numbers} \mathrm{are} \mathrm{formed} \mathrm{using} 1,2,3,4,5.$

$\mathrm{A}=\mathrm{three} \mathrm{digit} \mathrm{numbers} \mathrm{divisible} \mathrm{by} 5$

$$\begin{gathered} \mathrm{keeping} 5 \mathrm{in} \mathrm{the} \mathrm{last} \mathrm{place} \mathrm{first} 2 \mathrm{places} \\ \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{using} 1,2,3,4 \mathrm{in} {}^4\mathrm{P}_2 \mathrm{ways}. \\ \therefore \mathrm{n}\left(\mathrm{A}\right)=4{\mathrm{P}}_2=12 \end{gathered}$$

$$\begin{gathered} \mathrm{B}=\mathrm{numbers} \mathrm{divisible} \mathrm{by} 3 \\ \mathrm{using} \left\{(3,4,5) (1,3,5) (2,3,4) (1,2,3)\right\} \\ \mathrm{n}\left(\mathrm{B}\right)=4(3!)=24 \end{gathered}$$

$$\begin{gathered} \mathrm{A}\cap \mathrm{B}= \mathrm{numbers} \mathrm{divisible} \mathrm{by} 15 \\ \boxed{3}\boxed{4}\boxed{5} \boxed{4}\boxed{3}\boxed{5} \boxed{1}\boxed{3}\boxed{5} \boxed{3}\boxed{1}\boxed{5} \\ \mathrm{n}(\mathrm{A}\cap \mathrm{B})=4 \end{gathered}$$

$\begin{array}{rcl}\mathrm{n}(\mathrm{A}\cup \mathrm{B})& =& \mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})\\ & =& 12+24-4\\ & =& 32\end{array}$