IF `0 lt r lt s le n and ""^n P_r = "^n P_s` then the value of r+s is

Options

A. 1
B. 2
C. `2n-1`
D. `2n-2`

Answer

To solve the problem, we start with the given condition that \( ^nP_r = ^nP_s \).

### Step-by-Step Solution:

1. **Understanding the Permutation Formula**:
The formula for permutations is given by:
\[ ^nP_r = \frac{n!}{(n-r)!} \]
and
\[ ^nP_s = \frac{n!}{(n-s)!} \]

2. **Setting the Equations Equal**:
Since we know that \( ^nP_r = ^nP_s \), we can write:
\[ \frac{n!}{(n-r)!} = \frac{n!}{(n-s)!} \]

3. **Canceling \( n! \)**:
We can cancel \( n! \) from both sides (assuming \( n! \neq 0 \)):
\[ \frac{1}{(n-r)!} = \frac{1}{(n-s)!} \]

4. **Cross Multiplying**:
This leads to:
\[ (n-s)! = (n-r)! \]

5. **Analyzing Factorials**:
The equality \( (n-s)! = (n-r)! \) implies that \( n-s = n-r \) or \( n-s = 0 \) (since factorials are equal only when their arguments are equal or one of them is zero).

6. **Finding Relationships**:
- From \( n-s = n-r \), we have \( s = r \), which contradicts \( r < s \).
- Hence, we must have \( n-s = 0 \) leading to \( s = n \).

7. **Substituting \( s \)**:
Now substituting \( s = n \) into the inequality \( 0 < r < s \):
\[ 0 < r < n \]

8. **Finding \( r \)**:
Since \( s = n \), we can now find \( r \) using the factorial equality:
\[ (n-n)! = (n-r)! \implies 0! = (n-r)! \]
This means \( n-r = 0 \) or \( r = n \), which again contradicts \( r < s \).

9. **Conclusion**:
The only valid solution is to assume \( r = n-1 \) and \( s = n \). Thus:
\[ r + s = (n-1) + n = 2n - 1 \]

### Final Answer:
The value of \( r + s \) is \( 2n - 1 \).