Question

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 _________

Hint:

For divisibility by 3, first find all combinations of digits that sum to a multiple of 3, then calculate permutations for each.

Answer 1

Correct Answer: 32

Step 1: Total numbers between 100-1000 are 3-digit numbers. Total possible $= 5 \times 4 \times 3 = 60$.
Step 2: Divisible by 5: Last digit must be 5. Remaining 2 places filled by $\{1, 2, 3, 4\}$ in $4 \times 3 = 12$ ways.
Step 3: Divisible by 3: Sum of digits must be a multiple of 3. Sets: $\{1,2,3\}, \{1,3,5\}, \{2,3,4\}, \{3,4,5\}, \{1,2,x \text{ no}\}, \{1,5,x .......\}$. Valid sets: $\{1,2,3\}, \{1,3,5\}, \{2,3,4\}, \{3,4,5\}, \{1,5,x \text{ no}\}, \{2,4,x \text{ no}\}$. Actually, sets are: (1,2,3), (1,3,5), (2,3,4), (3,4,5), (1,5,? no), (4,5,? no). Sets summing to mult of 3: $\{1,2,3\}, \{1,3,5\}, \{2,3,4\}, \{3,4,5\}, \{1,5,3\}$ (repeated). Unique sets: $\{1,2,3\}, \{1,3,5\}, \{1,5,? \text{ no}\}, \{2,3,4\}, \{3,4,5\}$.
Step 4: Applying $n(A \cup B) = n(A) + n(B) - n(A \cap B)$, carefully counting perms gives 32.