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📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 Alternatively, each object can go into one of the two subsets, giving \(2^7 = 128\) labeled assignments, but dividing by 2 for indistinguishability gives \(64\), and subtracting the two cases where all are in one zone (leaving the other empty), we get \(64 - 1 = 63\) (we subtract 1 for the all-empty-one-case, but actually, each symmetric split is counted once, so the standard result applies directly).

📌 Thus, the number of distinct groupings is \(\boxed{63}\).

📌 Question:** A data analyst in Toronto is analyzing sequences of 8 diagnostic test results, each being either positive (P) or negative (N). How many such sequences contain exactly 3 positive results with no two consecutive positives?

📌 We are to count the number of binary sequences of length 8 with exactly 3 P's (positive) and 5 N's (negative), such that no two P's are adjacent.

📌 First, place the 5 N's. This creates 6 possible "gaps" where P's can be placed — one before the first N, one after each N (between N's), and one after the last N. For example:

📌 \(_ \, N \, _ \, N \, _ \, N \, _ \, N \, _ \, N \, _ \) → 6 gaps.

📌 We must place 3 P's into these 6 gaps, with at most one P per gap (to prevent adjacency).

📌 The number of ways to choose 3 distinct gaps from 6 is:

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