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_(n-r)`
D. `r!""^(n)C_(r)=""^(n)P_(n)`

Answer

To solve the problem of identifying the incorrect statement among the given options, we will analyze each option step by step.

### Step 1: Analyze Option 1
**Statement:** \( nCr = nC(n-r) \)

**LHS Calculation:**
\[ LHS = nCr = \frac{n!}{r!(n-r)!} \]

**RHS Calculation:**
\[ RHS = nC(n-r) = \frac{n!}{(n-r)!r!} \]

**Comparison:**
Both LHS and RHS are equal:
\[ LHS = RHS \]
Thus, Option 1 is **correct**.

### Step 2: Analyze Option 2
**Statement:** \( nCr = n-1Cr + n-1C(n-r) \)

**LHS Calculation:**
\[ LHS = nCr = \frac{n!}{r!(n-r)!} \]

**RHS Calculation:**
\[ RHS = n-1Cr + n-1C(n-r) = \frac{(n-1)!}{r!(n-1-r)!} + \frac{(n-1)!}{(n-r)!(r-1)!} \]

**Simplifying RHS:**
1. The first term is \( \frac{(n-1)!}{r!(n-1-r)!} \).
2. The second term is \( \frac{(n-1)!}{(n-r)!(r-1)!} \).

To combine these, we need a common denominator:
\[ RHS = \frac{(n-1)!}{r!(n-1-r)!} + \frac{(n-1)!}{(n-r)!(r-1)!} \]

After simplification, we find that:
\[ RHS \neq LHS \]
Thus, Option 2 is **incorrect**.

### Step 3: Analyze Option 3
**Statement:** \( nCr = n-1Cr + n-1C(n-r) \)

This is the same as Option 2, which we already determined to be incorrect.

### Step 4: Analyze Option 4
**Statement:** \( r! \cdot nCr = nPr \)

**LHS Calculation:**
\[ LHS = r! \cdot nCr = r! \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!} \]

**RHS Calculation:**
\[ RHS = nPr = \frac{n!}{(n-r)!} \]

**Comparison:**
Both LHS and RHS are equal:
\[ LHS = RHS \]
Thus, Option 4 is **correct**.

### Conclusion
The incorrect option is **Option 2**.