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Home » These 7 Good PS3 Games Will Blow Your Theater Experience! - American Beagle Club

These 7 Good PS3 Games Will Blow Your Theater Experience! - American Beagle Club

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Model: Place 3 ones such that no two are adjacent. To enforce non-adjacency, we can use the "stars and bars with gaps" method.

📌 Represent placing 3 ones with at least one zero between each. First place 3 ones with a zero between each: 1 0 1 0 1 → uses 5 positions, leaving $ 8 - 5 = 3 $ zeros to distribute freely in the 4 gaps: before first 1, between first and second, between second and third, and after third.

📌 Let the number of extra zeros in these 4 gaps be $ x_1, x_2, x_3, x_4 \geq 0 $, with sum $ x_1 + x_2 + x_3 + x_4 = 3 $

📌 Number of non-negative integer solutions: $ \binom{3 + 4 - 1}{3} = \binom{6}{3} = 20 $

📌 So there are 20 valid configurations (binary strings) with exactly 3 non-adjacent 1s.

📌 For each such configuration, the probability is:

📌 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^5 = \frac{1^3 \cdot 2^5}{3^8} = \frac{32}{6561}

📌 Since each configuration has the same probability and they are mutually exclusive:

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2z = 30^\circ + 360^\circ n \quad \text{or} \quad 2z = 330^\circ + 360...
Divide by 2 to find \(z\):
z = 15^\circ + 180^\circ n \quad \text{or} \quad z = 165^\circ + 180^\...
For \(z \in [0^\circ, 360^\circ]\), evaluate:
\(z = 15^\circ, 195^\circ\) from \(15^\circ + 180^\circ n\)
\(z = 165^\circ, 345^\circ\) from \(165^\circ + 180^\circ n\)
Sum these angles:
15^\circ + 195^\circ + 165^\circ + 345^\circ = 720^\circ
Thus, the sum of all angles is:
\boxed{720^\circ}

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