At an election meeting 10 speakers are to address the meeting. The only protocol to be observed is that whenever they speak PM will speak before MP and MP will speaks before MLA. The number of ways in which the meeting be addressed is…

Answer

To solve the problem of how many ways 10 speakers can address a meeting while adhering to the protocol that the PM must speak before the MP, and the MP must speak before the MLA, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Speakers**:
We have a total of 10 speakers.

2. **Understand the Protocol**:
The protocol states that:
- PM must speak before MP.
- MP must speak before MLA.
This means that in any arrangement of the speakers, the PM, MP, and MLA must appear in the order PM, MP, MLA.

3. **Calculate the Total Arrangements Without Protocol**:
If there were no restrictions, the total number of arrangements of 10 speakers would be given by the factorial of the number of speakers:
\[ \text{Total arrangements} = 10! \]

4. **Determine the Number of Ways to Arrange PM, MP, and MLA**:
The three specific roles (PM, MP, MLA) can be arranged in 3! (factorial of 3) ways. However, since we have a specific order (PM before MP before MLA), there is only 1 valid arrangement out of the 6 possible arrangements of these three speakers:
\[ \text{Valid arrangements of PM, MP, MLA} = 1 \]

5. **Calculate the Total Arrangements Considering the Protocol**:
Since the arrangement of PM, MP, and MLA is fixed, we can treat them as a single unit. This means we have 8 other speakers plus this unit (PM, MP, MLA), making a total of 8 + 1 = 9 units to arrange:
\[ \text{Total arrangements with protocol} = 8! \times 1 = 8! \]

6. **Final Calculation**:
Now we can compute the final answer:
\[ \text{Total arrangements with protocol} = \frac{10!}{3!} = \frac{10!}{6} = \frac{3628800}{6} = 604800 \]

### Final Answer:
The number of ways in which the meeting can be addressed, adhering to the protocol, is **604800**.