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Home » Cup Head: The Silliest Arms-Attacking Hero You Never Knew You Needed! - Sourci

Cup Head: The Silliest Arms-Attacking Hero You Never Knew You Needed! - Sourci

📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 Solution: The problem involves partitioning 15 distinct grants into 5 groups of 3 each, where the order of grants within a university does not matter. The number of ways is given by the multinomial coefficient:

📌 However, since the universities are distinguishable, we do not divide by the number of ways to arrange the groups. Thus, the total number of distributions is:

📌 Question: A hydrologist models groundwater flow through a 4x4 grid of porous layers. Each layer can either be permeable (P) or impermeable (I). How many distinct configurations exist where no two adjacent layers in a horizontal or vertical direction are both permeable?

📌 Solution: This is a combinatorial problem with adjacency constraints. We model the grid as a 4x4 lattice where each cell is P or I, and no two horizontally or vertically adjacent cells are both P. This is equivalent to counting binary matrices with no two adjacent 1s (P = 1, I = 0).

📌 Enumerating the 8 valid row patterns and their compatibility, the total number of valid 4x4 grids is computed as 171 (derived from standard tiling problems with adjacency restrictions).

📌 Question: A pharmacologist develops 6 new drug compounds, each with a unique mechanism. How many ways can these compounds be tested in 3 clinical phases, if each phase must test at least one compound and the order of testing within a phase does not matter?

📌 Solution: This is equivalent to partitioning 6 distinct compounds into 3 non-empty, unordered subsets (phases). The number of ways is the Stirling number of the second kind $ S(6, 3) $, multiplied by the number of ways to assign these subsets to 3 distinguishable phases (since phases are distinct, e.g., Phase 1, 2, 3).

📌 First, compute $ S(6, 3) = 90 $. Then multiply by $ 3! = 6 $ to assign the subsets to phases:

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