Chapter 2 - Section 2-1 - Sample Spaces and Events - Exercises - Page 28: 2-42

(a) 95,040 (b) 792

(a) If the chips are of different types, then every arrangement of 5 locations selected from the 12 results in a different layout. Therefore, $12P5 = \frac{12!}{(12-5)!} = 95,040$ (b) If the chips are of the same type, then every subset of 5 locations chosen from the 12 results in a different layout. Therefore, $12C5 = \frac{12!}{5! 7!} = 792$