Discrete Maths - Combinatorics

16. Let P (n) : 2n < n!, for all n ≥ 4. Then P (6) is

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17. For all n ∈ N, 4n − 1 is always divisible by

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18. Let A = {1, 2, 3, ..., k}. What should be the minimum value of k such that always there are at least 2 numbers having same remainder when divided by 5?

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