<p>There are unlimited number of identical balls of four different colours. How many arrangements of at most 8 balls in a row can be made by using them?</p>
Step-by-Step Solution
Key Concept: Each of the at most 8 positions can independently be filled with any of 4 colours, so we sum the arrangements for 1 ball through 8 balls using the multiplication principle: total = 4¹ + 4² + ... + 4⁸.
<p><strong>Step 1:</strong> Identify the problem structure. We need arrangements of 1 ball, 2 balls, 3 balls, ..., up to 8 balls in a row, where each ball can be any of 4 colours.</p><p><strong>Step 2:</strong> For k balls in a row, each position can be filled with 4 different colours independently. Number of arrangements with k balls = 4<sup>k</sup></p><p><strong>Step 3:</strong> Total arrangements = 4¹ + 4² + 4³ + ... + 4⁸</p><p><strong>Step 4:</strong> This is a geometric series with first term a = 4, common ratio r = 4, and n = 8 terms.</p><p><strong>Step 5:</strong> Using the formula S<sub>n</sub> = a(r<sup>n</sup> - 1)/(r - 1):</p><p>S = 4(4⁸ - 1)/(4 - 1) = 4(65536 - 1)/3 = 4(65535)/3 = 262140/3 = 87380</p><p>∴ Answer: <strong>87380</strong></p>
Correct Answer: 87380