There are m seats in the first row of a theatre, of which n are to be occupied.The number of ways of arranging n persons if no two persons sit side by side is

Options

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

Answer

To solve the problem of arranging \( n \) persons in \( m \) seats such that no two persons sit side by side, we can follow these steps:

### Step 1: Understand the arrangement
We have \( m \) seats and we need to arrange \( n \) persons in such a way that no two persons are adjacent. This means that between any two persons, there must be at least one empty seat.

### Step 2: Calculate the minimum seats required
To seat \( n \) persons with at least one empty seat between each pair, we need:
- \( n \) persons
- \( n-1 \) empty seats (to separate the \( n \) persons)

Thus, the minimum number of seats required is:
\[ n + (n - 1) = 2n - 1 \]

### Step 3: Check if arrangement is possible
For the arrangement to be possible, we must have:
\[ m \geq 2n - 1 \]
If \( m < 2n - 1 \), it is impossible to arrange the persons as required.

### Step 4: Calculate the remaining seats
If \( m \geq 2n - 1 \), we can calculate the number of remaining seats after placing \( n \) persons and \( n-1 \) empty seats:
\[ \text{Remaining seats} = m - (2n - 1) = m - 2n + 1 \]

### Step 5: Distribute the remaining seats
Now, we have \( m - 2n + 1 \) remaining seats that can be distributed freely. We can think of these remaining seats as additional empty spaces that can be placed before the first person, between persons, or after the last person.

### Step 6: Formulate the problem
Let \( x_1 \) be the number of empty seats before the first person, \( x_2 \) be the number of empty seats between the first and second persons, and so on up to \( x_{n+1} \) which is the number of empty seats after the last person. The equation becomes:
\[ x_1 + x_2 + x_3 + ... + x_{n+1} = m - 2n + 1 \]
where \( x_2, x_3, ..., x_n \) must be at least 0 (since there can be no restriction on the number of empty seats in these positions), and \( x_1, x_{n+1} \) can also be 0.

### Step 7: Use the stars and bars method
The number of non-negative integer solutions to the equation \( x_1 + x_2 + ... + x_{n+1} = m - 2n + 1 \) can be found using the stars and bars combinatorial method. The formula for the number of ways to distribute \( r \) indistinguishable objects (remaining seats) into \( k \) distinguishable boxes (the gaps between and around the persons) is given by:
\[ \binom{r + k - 1}{k - 1} \]
In our case, \( r = m - 2n + 1 \) and \( k = n + 1 \):
\[ \text{Number of ways} = \binom{(m - 2n + 1) + (n + 1) - 1}{(n + 1) - 1} = \binom{m - n + 1}{n} \]

### Step 8: Arrange the persons
Since the \( n \) persons are distinct, we can arrange them in \( n! \) ways.

### Step 9: Final answer
The total number of arrangements is:
\[ \text{Total arrangements} = \binom{m - n + 1}{n} \times n! \]

### Conclusion
Thus, the number of ways of arranging \( n \) persons in \( m \) seats such that no two persons sit side by side is:
\[ \boxed{\binom{m - n + 1}{n} \times n!} \]