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Home » Why Every Madea Movie and TV Show is a Must-Watch: Spoiler Alerts Included! - American Beagle Club

Why Every Madea Movie and TV Show is a Must-Watch: Spoiler Alerts Included! - American Beagle Club

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 We attempt to **count valid sequences by placing high-frequency letters first with separation**, then fill with others.

📌 Let’s suppose we place the **C’s** with no two adjacent.

📌 To place 3 C’s with no two adjacent in 8 positions: we first place 3 C’s such that there is at least one space between them.

📌 The number of ways to choose 3 non-adjacent positions in 8 is a classic stars and bars adjustment.

📌 To place 3 non-consecutive positions: let the positions be $ x_1 < x_2 < x_3 $, with $ x_{i+1} \ge x_i + 2 $. Let $ y_i = x_i - (i-1) $, then $ y_1 < y_2 < y_3 $ in $ \{1,\dots,6\} $. So number is $ \binom{6}{3} = 20 $.

📌 For each such choice of 3 non-adjacent positions, we place 3 non-adjacent C’s.

📌 Now, from the remaining 5 positions, we must place 2 A’s and 3 G’s (since 3 C’s are placed), **with no two A’s adjacent and no two G’s adjacent** — but wait: the condition is **no two identical letters adjacent**, so A’s cannot be adjacent to A, G’s to G.

📌 But A and G are different, so A and G can be adjacent.

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Solutions: \( x = 3 \) or \( x = 1 \).
#### 1, 3
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