Which of the following is incorrect.

Options

A. `""^n C_r="^n C_(n-r)`
B. `""^n C_r=""^(n-1) C_r+"^n C_(n-r)`
C. `""^n C_r=""^(n-1) C_r+"^(n-1) C_(r-1)`
D. `r!""^n C_r="^n P_r`

Answer

To determine which of the given options is incorrect, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Option A: \( nCr = nC(n-r) \)**
- **LHS:** \( nCr = \frac{n!}{r!(n-r)!} \)
- **RHS:** \( nC(n-r) = \frac{n!}{(n-r)!r!} \)
- Both sides are equal.
- **Conclusion:** Option A is correct.

2. **Option B: \( nCr = n-1Cr + nC(n-r) \)**
- **LHS:** \( nCr = \frac{n!}{r!(n-r)!} \)
- **RHS:**
- First term: \( n-1Cr = \frac{(n-1)!}{r!(n-1-r)!} \)
- Second term: \( nC(n-r) = \frac{n!}{(n-r)!r!} \)
- Combining these, we have:
\[ n-1Cr + nC(n-r) = \frac{(n-1)!}{r!(n-1-r)!} + \frac{n!}{(n-r)!r!} \]
- This does not simplify to \( nCr \).
- **Conclusion:** Option B is incorrect.

3. **Option C: \( nCr = n-1Cr + n-1C(r-1) \)**
- **LHS:** \( nCr = \frac{n!}{r!(n-r)!} \)
- **RHS:**
- First term: \( n-1Cr = \frac{(n-1)!}{r!(n-1-r)!} \)
- Second term: \( n-1C(r-1) = \frac{(n-1)!}{(r-1)!(n-r)!} \)
- Combining these gives:
\[ n-1Cr + n-1C(r-1) = \frac{(n-1)!}{r!(n-1-r)!} + \frac{(n-1)!}{(r-1)!(n-r)!} \]
- This simplifies to \( nCr \).
- **Conclusion:** Option C is correct.

4. **Option D: \( r! \cdot nCr = nPr \)**
- **LHS:** \( r! \cdot nCr = r! \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!} \)
- **RHS:** \( nPr = \frac{n!}{(n-r)!} \)
- Both sides are equal.
- **Conclusion:** Option D is correct.

### Final Conclusion:
The incorrect option is **Option B**.

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