VTU 4th Sem Math | Counting Numbers Divisible by 5, 6, or 8 | Principles of Counting II - Module 4
n this video, I solve an important problem from Principles of Counting II, Module 4 of the 4th Semester VTU Mathematics syllabus. The question is: How many integers between 1 and 300 (inclusive) are: (i) Divisible by at least one of 5, 6, or 8? (ii) Divisible by none of 5, 6, or 8? This problem is explained step by step using the principle of inclusion-exclusion, making it easy for VTU students to understand how to efficiently count numbers with multiple divisibility conditions.

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Find Numbers Avoiding 5, 6, and 8 Between 1–300 | Inclusion-Exclusion Principle

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VTU 4th Sem Math | Counting Numbers Not Divisible by 2, 3, or 5 | Principles of Counting - Module 4

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Find Arrangements Without CAR, DOG, PUN, BYTE | Discrete Math Exam Question | VTU

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VTU 4th Sem Math | Solve Recurrence Relation | Module 4 | Cn = 3Cn-1 – 2Cn-2 | Recurrence Formula

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Sum of an AP Class 10 Maths Chapter 5 | Arithmetic Progression Formulas & Questions

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How to Calculate Faster than a Calculator - Mental Math #1

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Who is Smarter? Engineer vs Chinese 5th Grader
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PINK & ORANGE GRADIENT IN HD [3 HOURS]

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Principle of Inclusion & Exclusion with Examples || Inclusion and Exclusion Problems || DMS

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