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

Answer

To solve the problem where \( \binom{16}{r} = \binom{16}{r+6} \), we can follow these steps:

### Step 1: Understand the property of binomial coefficients
We know that:
\[ \binom{n}{r} = \binom{n}{n-r} \]
This means that \( \binom{16}{r} = \binom{16}{16-r} \).

### Step 2: Set up the equation
From the given equation:
\[ \binom{16}{r} = \binom{16}{r+6} \]
Using the property of binomial coefficients, we can rewrite the right-hand side:
\[ \binom{16}{r+6} = \binom{16}{16 - (r+6)} = \binom{16}{10 - r} \]
Thus, we have:
\[ \binom{16}{r} = \binom{16}{10 - r} \]

### Step 3: Equate the indices
From the equality of the binomial coefficients, we can equate the indices:
\[ r = 10 - r \]
This simplifies to:
\[ 2r = 10 \]
So,
\[ r = 5 \]

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

### Step 5: Calculate \( \binom{5}{5} \)
Using the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
We can substitute \( n = 5 \) and \( r = 5 \):
\[ \binom{5}{5} = \frac{5!}{5! \cdot (5-5)!} = \frac{5!}{5! \cdot 0!} \]
Since \( 0! = 1 \):
\[ \binom{5}{5} = \frac{5!}{5! \cdot 1} = 1 \]

### Final Answer
Thus, the value of \( \binom{5}{r} \) is:
\[ \boxed{1} \]