If `.^(18)C_(r) = .^(18)C_(r+2)`, then evaluate `.^(r)C_(5)`.

Answer

To solve the problem where \( ^{18}C_r = ^{18}C_{r+2} \) and we need to evaluate \( ^rC_5 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Combination Equality**:
We know that if \( ^nC_r = ^nC_k \), then either \( r = k \) or \( r + k = n \). In our case, we have:
\[ ^{18}C_r = ^{18}C_{r+2} \]
This implies:
\[ r + (r + 2) = 18 \]

2. **Setting Up the Equation**:
From the equation above, we can simplify:
\[ 2r + 2 = 18 \]

3. **Solving for \( r \)**:
Now, we can solve for \( r \):
\[ 2r = 18 - 2 \]
\[ 2r = 16 \]
\[ r = \frac{16}{2} = 8 \]

4. **Finding \( ^rC_5 \)**:
Now that we have \( r = 8 \), we need to evaluate \( ^8C_5 \):
\[ ^8C_5 = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \cdot 3!} \]

5. **Calculating the Factorials**:
We can express \( 8! \) as:
\[ 8! = 8 \times 7 \times 6 \times 5! \]
Thus, substituting this into our combination formula:
\[ ^8C_5 = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3!} \]
The \( 5! \) cancels out:
\[ ^8C_5 = \frac{8 \times 7 \times 6}{3!} \]

6. **Calculating \( 3! \)**:
We know that \( 3! = 3 \times 2 \times 1 = 6 \). Therefore:
\[ ^8C_5 = \frac{8 \times 7 \times 6}{6} \]

7. **Final Calculation**:
The \( 6 \) in the numerator and denominator cancels out:
\[ ^8C_5 = 8 \times 7 = 56 \]

### Final Answer:
Thus, the value of \( ^rC_5 \) is \( 56 \).