seven delegates are to address a meeting. If a particular peaker is to speak before another particular speaker, find the number of ways in which this can be arranged.

Options

A. 1220
B. 2520
C. 3250
D. 7826

Answer

To solve the problem of arranging seven delegates where a particular speaker (let's call them A) must speak before another particular speaker (let's call them B), we can follow these steps:

### Step-by-Step Solution:

1. **Total Delegates**: We have 7 delegates in total.

2. **Total Arrangements Without Restrictions**: If there were no restrictions, the total number of ways to arrange 7 delegates would be calculated using factorial notation:
\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]

3. **Considering the Condition (A must speak before B)**: Since we need A to speak before B, we can think about the arrangements in pairs. In any arrangement of A and B, there are two possibilities: A can be before B or B can be before A. Since we only want the arrangements where A is before B, we can divide the total arrangements by 2:
\[ \text{Arrangements with A before B} = \frac{7!}{2} = \frac{5040}{2} = 2520 \]

4. **Final Answer**: Therefore, the number of ways in which the delegates can be arranged such that A speaks before B is:
\[ \text{Number of arrangements} = 2520 \]

### Conclusion:
The final answer is **2520**.