Number of ways 10 speakers A1 to A10 can speak when A1 must always speak after A2

Introduction / Context:
This question is a classic example of permutations with an order restriction between two particular elements. You are given 10 speakers and a condition that one specific speaker must speak after another specific speaker. Such problems help build strong intuition for symmetry and counting methods.

Given Data / Assumptions:

• Speakers are labeled A1, A2, A3, up to A10.
• All speakers are distinct.
• All 10 speakers must speak exactly once.
• A1 must speak after A2 in the final order.
• No other restrictions are imposed.

Concept / Approach:
When there is a simple condition like A1 must come after A2 and there are no other constraints, we can use a symmetry argument. In all possible permutations of 10 distinct speakers, A1 and A2 are equally likely to appear in any relative order. Therefore exactly half of all permutations will satisfy A1 after A2 and the other half will have A1 before A2.

Step-by-Step Solution:
Step 1: Count total unrestricted permutations of 10 distinct speakers.Step 2: Total permutations = 10!.Step 3: Consider the pair (A1, A2). In any particular permutation, either A1 comes before A2 or A2 comes before A1.Step 4: There is perfect symmetry between these two possibilities, so the number of permutations where A1 is before A2 equals the number where A1 is after A2.Step 5: Therefore, the number of permutations with A1 after A2 is exactly half of all permutations.Step 6: Required count = 10! / 2.

Verification / Alternative check:
Another way is to think of first arranging all 10 speakers in 10! ways, then looking at each arrangement in pairs. If you swap the positions of A1 and A2 in any arrangement where A1 is before A2, you get a unique arrangement where A1 is after A2, and vice versa. This one to one correspondence confirms that the count is exactly half of 10!.

Why Other Options Are Wrong:

• 9! and 9!/2 ignore that all 10 speakers are being permuted, not 9.
• 10! counts all permutations without applying the restriction that A1 must be after A2.

Common Pitfalls:
Students sometimes attempt to count positions directly, such as choosing a position for A2 and later choosing a position for A1, which often leads to double counting or missing cases. The symmetry argument is both simpler and more reliable for this type of relative order condition.

Final Answer:
The number of valid speaking orders is 10! divided by 2, so the correct answer is 10! / 2.