For `2 le r le n , {:((n),(r):})+2{:((n),(r-1):})+{:((n),(r-2):})=`

Options

A. `{:((n+1),(r-1):})`
B. `2{:((n+1),(r+1):})`
C. `2{:((n+2),(r):})`
D. `{:((n+2),(r):})`

Answer

To solve the equation given in the question, we will use the properties of combinations (denoted as nCr) and the identity involving combinations.

### Step-by-Step Solution:

1. **Understand the Combination Notation**:
The notation \( nCr \) represents the number of ways to choose \( r \) items from \( n \) items without regard to the order of selection. It is calculated using the formula:
\[ nCr = \frac{n!}{r!(n-r)!} \]

2. **Identify the Given Expression**:
We need to simplify the expression:
\[ nCr + 2 \cdot nC(r-1) + nC(r-2) \]

3. **Apply the Combination Identity**:
We can use the identity:
\[ nCr + nC(r-1) = (r+1)nC(r) \]
This identity can be derived from the combinatorial interpretation of choosing items.

4. **Rewrite the Expression**:
Using the identity, we can rewrite the expression:
\[ nCr + 2 \cdot nC(r-1) + nC(r-2) = nCr + nC(r-1) + nC(r-1) + nC(r-2) \]
Now, applying the identity:
\[ = (r+1)nC(r) + nC(r-2) \]

5. **Combine the Terms**:
The expression now becomes:
\[ (r+1)nC(r) + nC(r-2) \]
We can further simplify this if needed, but generally, this is the simplified form of the expression.

### Final Result:
The expression simplifies to:
\[ nCr + 2 \cdot nC(r-1) + nC(r-2) = (r+1)nC(r) + nC(r-2) \]