Which of the following is correct

Options

A. `""^n P_n=2 "^n P_(n-2)`
B. `""^n P_n= "^n P_(n-1)`
C. `""^n P_r=""^(n-1) P_r+r"^(n-1) P_(r-1)`
D. `""^n P_r=n "^(n-1) P_(r-1)`

Answer

To determine which of the given options is correct, we will analyze each option step by step using the concept of permutations.

### Step-by-Step Solution:

**Option 1: \( nP_n = 2nP_{n-2} \)**

1. **Left Hand Side (LHS)**:
\[ nP_n = \frac{n!}{(n-n)!} = \frac{n!}{0!} = n! \]
(Since \(0! = 1\))

2. **Right Hand Side (RHS)**:
\[ 2nP_{n-2} = 2 \cdot \frac{n!}{(n-(n-2))!} = 2 \cdot \frac{n!}{2!} = 2 \cdot \frac{n!}{2} = n! \]

3. **Conclusion**:
\[ LHS = RHS \implies nP_n = 2nP_{n-2} \text{ is correct.} \]

---

**Option 2: \( nP_n = nP_{n-1} \)**

1. **Left Hand Side (LHS)**:
\[ nP_n = n! \]

2. **Right Hand Side (RHS)**:
\[ nP_{n-1} = \frac{n!}{(n-(n-1))!} = \frac{n!}{1!} = n! \]

3. **Conclusion**:
\[ LHS = RHS \implies nP_n = nP_{n-1} \text{ is correct.} \]

---

**Option 3: \( nP_r = (n-1)P_r + r(n-1)P_{r-1} \)**

1. **Left Hand Side (LHS)**:
\[ nP_r = \frac{n!}{(n-r)!} \]

2. **Right Hand Side (RHS)**:
\[ (n-1)P_r + r(n-1)P_{r-1} = \frac{(n-1)!}{(n-1-r)!} + r \cdot \frac{(n-1)!}{(n-1-(r-1))!} \]
Simplifying RHS:
\[ = \frac{(n-1)!}{(n-1-r)!} + r \cdot \frac{(n-1)!}{(n-r)!} \]
Factoring out \( (n-1)! \):
\[ = (n-1)! \left( \frac{1}{(n-1-r)!} + \frac{r}{(n-r)!} \right) \]
Finding a common denominator gives:
\[ = (n-1)! \cdot \frac{(n-r) + r}{(n-r)!} = (n-1)! \cdot \frac{n}{(n-r)!} = \frac{n!}{(n-r)!} \]

3. **Conclusion**:
\[ LHS = RHS \implies nP_r = (n-1)P_r + r(n-1)P_{r-1} \text{ is correct.} \]

---

**Option 4: \( nP_r = (n-1)P_{r-1} \)**

1. **Left Hand Side (LHS)**:
\[ nP_r = \frac{n!}{(n-r)!} \]

2. **Right Hand Side (RHS)**:
\[ (n-1)P_{r-1} = \frac{(n-1)!}{((n-1)-(r-1))!} = \frac{(n-1)!}{(n-r)!} \]

3. **Conclusion**:
\[ LHS \neq RHS \implies nP_r \neq (n-1)P_{r-1} \text{ is incorrect.} \]

### Final Conclusion:
- **Correct Options**: 1, 2, and 3 are correct.
- **Incorrect Option**: 4 is incorrect.