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📅 May 3, 2026 👤 scraface

May 03, 2026

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\( n = 4 \): 6−4=2 → \(\binom{2}{2} = 1\)
\( n = 5 \): 6−5=1 → \(\binom{1}{2} = 0\)
So only \( n = 4 \): 1 way
T is in the remaining position: once R’s and L’s are placed, T is fixe...
So total for \( m=3, n=4 \): 1 (R choice) × 1 (L choice) = 1
L positions: {4,5}, R: {1,2} → min L = 4, max R = 2 → 2 < 4 → satisfie...
So one valid sequence: R R L L T (but types indistinct)
But we don’t need to reconstruct — just count the number of full assig...
So far: 1
Case \( m = 4 \):

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