In an electrion the coinvencing will be done by 20, 25 and 30 persons in three districts. If 75 persons are avilable then in how many ways then can be selected?

Answer

To solve the problem of selecting persons for convincing in three districts, we need to follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Persons Available**:
We have a total of 75 persons available for selection.

2. **Determine the Number of Persons to be Selected for Each District**:
- For the first district, we need to select 20 persons.
- For the second district, we need to select 25 persons.
- For the third district, we need to select 30 persons.

3. **Calculate the Number of Ways to Select Persons**:
- First, we choose 20 persons from the 75 available. This can be represented as \( \binom{75}{20} \).
- After selecting 20 persons, we have 55 persons left. Now, we need to select 25 persons from these 55. This can be represented as \( \binom{55}{25} \).
- After selecting 25 persons, we have 30 persons left. We need to select all 30 persons from these remaining 30. This can be represented as \( \binom{30}{30} \).

4. **Combine the Selections**:
The total number of ways to select the persons is given by the product of the combinations calculated in the previous step:
\[ \text{Total Ways} = \binom{75}{20} \times \binom{55}{25} \times \binom{30}{30} \]

5. **Simplify the Expression**:
The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Therefore, we can express the combinations as:
\[ \binom{75}{20} = \frac{75!}{20! \cdot 55!} \]
\[ \binom{55}{25} = \frac{55!}{25! \cdot 30!} \]
\[ \binom{30}{30} = 1 \]
Thus, the total number of ways can be simplified to:
\[ \text{Total Ways} = \frac{75!}{20! \cdot 55!} \times \frac{55!}{25! \cdot 30!} \times 1 \]
The \( 55! \) cancels out:
\[ \text{Total Ways} = \frac{75!}{20! \cdot 25! \cdot 30!} \]

6. **Final Answer**:
The total number of ways to select the persons for the three districts is:
\[ \frac{75!}{20! \cdot 25! \cdot 30!} \]