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 ______

We need to form three-digit numbers (between 100 and 1000) using the digits $$\{1, 2, 3, 4, 5\}$$ without repetition, such that the number is divisible by either 3 or 5.

Using the inclusion-exclusion principle: the count of numbers divisible by 3 or 5 equals (divisible by 3) + (divisible by 5) - (divisible by 15).

A number is divisible by 3 if and only if the sum of its digits is divisible by 3. We list all 3-element subsets of $$\{1, 2, 3, 4, 5\}$$ whose digit sum is divisible by 3: $$\{1, 2, 3\}$$ with sum 6, $$\{1, 3, 5\}$$ with sum 9, $$\{2, 3, 4\}$$ with sum 9, and $$\{3, 4, 5\}$$ with sum 12. Each set of 3 digits can be arranged in $$3! = 6$$ ways, giving $$4 \times 6 = 24$$ numbers divisible by 3.

A number is divisible by 5 if and only if its units digit is 5. The remaining two positions are filled by choosing 2 digits from $$\{1, 2, 3, 4\}$$ and arranging them, giving $$4 \times 3 = 12$$ numbers divisible by 5.

A number is divisible by 15 if it is divisible by both 3 and 5. The units digit must be 5, and the digit sum must be divisible by 3. We need two digits from $$\{1, 2, 3, 4\}$$ such that their sum plus 5 is divisible by 3. Checking all pairs: $$\{1, 3\}$$ gives sum 9 (yes), and $$\{3, 4\}$$ gives sum 12 (yes). The remaining pairs $$\{1, 2\}, \{1, 4\}, \{2, 3\}, \{2, 4\}$$ give sums 8, 10, 10, 11 respectively (none divisible by 3). Each valid pair can be arranged in $$2! = 2$$ ways in the hundreds and tens places, giving $$2 \times 2 = 4$$ numbers divisible by 15.

By inclusion-exclusion, the total count is $$24 + 12 - 4 = 32$$.