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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
A five digit number divisible by 3 using digits 0, 1, 2, 3, 4, 5 is to be made
without repetition. Find of such word.
Solution
Digits available: {0,1,2,3,4,5}.
Sum of all six = 15 divisible by 3.
To make a 5-digit number divisible by 3, exclude one digit whose value 0, i.e., exclude 0 or 3.
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Exclude 0 → use {1,2,3,4,5}: all 5! = 120 numbers (no leading-zero issue).
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Exclude 3 → use {0,1,2,4,5}: total 5! − 4! (leading 0) = 120 − 24 = 96.
Total = 120 + 96 = 216
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