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Home » Counting all combinations for \((a, b)\): - ECD Germany

Counting all combinations for \((a, b)\): - ECD Germany

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 For each such pair (m, k), we count the number of ways:

📌 Choose 2 distinct positions for R’s: both < m? No: R’s must be all \( < k \), and max \( R \leq m \), and \( m < k \), so all R positions ∈ [1, m-1]? No: if \( k \geq m+1 \), we need R’s all < k, but \( m < k \), so R’s can be in [1, k−1], but they must all be < k and their max < k, but we require max R < k — always true if R’s are in [1, k−1]. But we need **both R’s < k**, which is automatic if we choose from 1 to k−1, but the condition is only upper bound on L’s.

📌 Wait — correction: we need **all R’s before all L’s**, i.e., every R position < every L position.

📌 So: max(R) < min(L)

📌 Let \( m = \max(\text{R positions}) \), \( n = \min(\text{L positions}) \), require \( m < n \)

📌 Now, \( m \geq 2 \), \( n \leq 5 - m \), but \( n > m \), and \( n \geq m+1 \)

📌 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.

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