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jim irsay american primeval - Moon Smoking

📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 Total number of 1×1 grid positions: \( 4 \times 3 = 12 \).

📌 Number of ways to choose 3 distinct positions: \( \binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220 \).

📌 Now, favorable outcomes: all 3 positions in different rows.

📌 There are 4 rows, choose 3 distinct rows: \( \binom{4}{3} = 4 \).

📌 Each chosen row has 3 columns, so for each row, pick 1 position: \( 3 \times 3 \times 3 = 27 \).

📌 But this counts selections per row order — since we are selecting unordered triples, we must divide by \( 3! = 6 \) only if order doesn’t matter. However, since we are selecting positions and the grid is fixed, each position is unique, and choosing one per row from 3 distinct rows gives \( 3^3 = 27 \) per set of 3 rows.

📌 So, number of favorable selections:

📌 \binom{4}{3} \times 3^3 = 4 \times 27 = 108

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