If `.^(16)C_(r) = .^(16)C_(r+6)`, then find `.^(5)C_(r)`.

Answer

To solve the problem, we start with the equation given in the question:

\[ \binom{16}{r} = \binom{16}{r+6} \]

### Step 1: Use the property of combinations
We know that the property of combinations states:

\[ \binom{n}{k} = \binom{n}{n-k} \]

Applying this property to our equation, we can rewrite \(\binom{16}{r+6}\) as:

\[ \binom{16}{r+6} = \binom{16}{16 - (r + 6)} = \binom{16}{10 - r} \]

### Step 2: Set the equations equal
Now we have:

\[ \binom{16}{r} = \binom{16}{10 - r} \]

### Step 3: Use the property of equality
For the above equation to hold true, we can deduce that:

\[ r = 10 - r \quad \text{or} \quad r = r + 6 \text{ (which is not possible)} \]

From \(r = 10 - r\):

\[ 2r = 10 \quad \Rightarrow \quad r = 5 \]

### Step 4: Find \(\binom{5}{r}\)
Now that we have \(r = 5\), we need to find \(\binom{5}{r}\):

\[ \binom{5}{5} = 1 \]

### Final Answer
Thus, the value of \(\binom{5}{r}\) is:

\[ \boxed{1} \]