If eight persons are to address a meeting, then the number of ways in which a specified speaker is to speak before another specified speaker, is

Options

A. 2520
B. 20160
C. 40320
D. none of these

Answer

To solve the problem of how many ways a specified speaker can speak before another specified speaker among eight persons, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Speakers**:
We have a total of 8 persons who will be addressing the meeting.

2. **Calculate the Total Arrangements**:
The total number of ways to arrange 8 persons is given by the factorial of 8, which is denoted as \(8!\).
\[ 8! = 40320 \]

3. **Consider the Condition**:
We need to ensure that a specified speaker (let's call them M1) speaks before another specified speaker (M2).

4. **Divide the Arrangements**:
In any arrangement of the 8 speakers, M1 can either speak before M2 or after M2. Since these two scenarios are equally likely, we can divide the total arrangements by 2 to find the number of arrangements where M1 speaks before M2.
\[ \text{Number of ways M1 speaks before M2} = \frac{8!}{2} = \frac{40320}{2} = 20160 \]

5. **Final Answer**:
Therefore, the number of ways in which the specified speaker M1 can speak before the specified speaker M2 is:
\[ \text{Final Answer} = 20160 \]