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Home » At d=20: N(20) = k / √2 = 450 → k = 450√2 - Groen Casting

At d=20: N(20) = k / √2 = 450 → k = 450√2 - Groen Casting

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 So for each \( m = 2,3,4 \), and \( n = m+1, m+2, \dots, 5 \), we count:

📌 Number of ways to choose 2 distinct positions for R’s from those \( < m \)? No: we need R’s chosen from positions 1 to \( m-1 \), but no: since max R must be < m, so R’s ∈ {1,2,...,m−1}, and we choose 2 of them: number is \(\binom{m-1}{2}\) if \( m-1 \geq 2 \), else 0.

📌 Number of ways to choose 2 distinct positions for L’s from positions \( \geq n \), and \( n \leq 5 \), so from {n, n+1, ..., 5}, which has size \( 5 - n + 1 \), and we choose 2: \(\binom{6 - n}{2}\), provided \( 6 - n \geq 2 \), i.e., \( n \leq 4 \)

📌 And T goes to the remaining position.

📌 Now compute for each \( m \):

📌 Case \( m = 2 \):

📌 R’s must be in {1}, but need 2 indistinct positions → need 2 positions < 2 → only position 1 → only one position → cannot choose 2 → \(\binom{1}{2} = 0\)

📌 So 0 ways.

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