JEE PYQ: Permutation And Combination Question 10
Question 10 - 2021 (25 Feb Shift 1)
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
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Answer: 32
Solution
Numbers divisible by 3: sum of digits divisible by 3. Possibilities: $(1,2,3), (3,4,5), (1,5,3), (2,3,4)$ giving $4 \times 3! = 24$. Numbers divisible by 5 (ending in 5): $4 \times 3 = 12$. Overlap (divisible by both 3 and 5): numbers ending in 5 with digit sum divisible by 3: $12 \to 3! = 6, 15 \to 3! = 6, 24 \to 3! = 6, 42 \to 3! = 6$… Inclusion-exclusion gives $24 + 12 - 4 = 32$.
BETA
