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.

February 22, 2026

$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 $, else 0.$

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 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 $:

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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