There are n number of seats and m number of people have to be seated, then how many ways are possible to do this `(mltn)`?

Options

A. `.^(n)P_(m)`
B. `.^(n)C_(m)`
C. `.^(n)C_(n)xx(m-1)!`
D. `.^(n-1)P_(m-1)`

Answer

To solve the problem of seating \( m \) people in \( n \) seats, we can break it down into two main steps: selecting the seats and arranging the people in those seats.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have \( n \) seats and \( m \) people, where \( n > m \). We need to find out how many ways we can seat these \( m \) people in the \( n \) seats.

2. **Selecting Seats**:
First, we need to select \( m \) seats from the available \( n \) seats. The number of ways to choose \( m \) seats from \( n \) seats can be calculated using the combination formula:
\[ \binom{n}{m} = \frac{n!}{m!(n-m)!} \]
This represents the number of ways to choose \( m \) seats from \( n \).

3. **Arranging People**:
After selecting the \( m \) seats, we need to arrange \( m \) people in those \( m \) seats. The number of ways to arrange \( m \) people is given by \( m! \) (factorial of \( m \)).

4. **Total Arrangements**:
To find the total number of ways to seat \( m \) people in \( n \) seats, we multiply the number of ways to choose the seats by the number of ways to arrange the people:
\[ \text{Total Ways} = \binom{n}{m} \times m! = \frac{n!}{m!(n-m)!} \times m! \]
The \( m! \) in the numerator and denominator cancels out:
\[ \text{Total Ways} = \frac{n!}{(n-m)!} \]

5. **Using Permutation Notation**:
The expression \( \frac{n!}{(n-m)!} \) can also be expressed using permutation notation as \( nPm \), which represents the number of ways to arrange \( m \) people in \( n \) seats.

### Final Answer:
Thus, the total number of ways to seat \( m \) people in \( n \) seats is:
\[ nPm = \frac{n!}{(n-m)!} \]