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Five speakers A, B, C, D and E are to speak at a meeting. How many different orders are possible if B is not allowed to come before A, whether directly or at any earlier position?
- 120
- 24
- 60
- 5⁴ × 4
Correct answer: 60
Solution
By symmetry, in half of all 5!=120 arrangements A precedes B and in half B precedes A. Requiring A before B gives 120/2 = 60 orders.
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