If `f (n) = sum_(s=1)^n sum_(r=s)^n "^nC_r`` "^rC_s` , then `f(3) =`

Options

A. 27
B. 19
C. 1
D. 5

Explanation

To solve the problem, we need to compute the function \( f(n) \) defined as:

\[ f(n) = \sum_{s=1}^{n} \sum_{r=s}^{n} \binom{n}{r} \binom{r}{s} \]

We are tasked with finding \( f(3) \).

### Step-by-Step Solution

1. **Set Up the Summation for \( n = 3 \)**:
\[ f(3) = \sum_{s=1}^{3} \sum_{r=s}^{3} \binom{3}{r} \binom{r}{s} \]

2. **Evaluate the Inner Summation for Each Value of \( s \)**:
- **Case 1: \( s = 1 \)**:
\[ \sum_{r=1}^{3} \binom{3}{r} \binom{r}{1} \]
- For \( r = 1 \): \( \binom{3}{1} \binom{1}{1} = 3 \cdot 1 = 3 \)
- For \( r = 2 \): \( \binom{3}{2} \binom{2}{1} = 3 \cdot 2 = 6 \)
- For \( r = 3 \): \( \binom{3}{3} \binom{3}{1} = 1 \cdot 3 = 3 \)
- Total for \( s = 1 \): \( 3 + 6 + 3 = 12 \)

- **Case 2: \( s = 2 \)**:
\[ \sum_{r=2}^{3} \binom{3}{r} \binom{r}{2} \]
- For \( r = 2 \): \( \binom{3}{2} \binom{2}{2} = 3 \cdot 1 = 3 \)
- For \( r = 3 \): \( \binom{3}{3} \binom{3}{2} = 1 \cdot 3 = 3 \)
- Total for \( s = 2 \): \( 3 + 3 = 6 \)

- **Case 3: \( s = 3 \)**:
\[ \sum_{r=3}^{3} \binom{3}{r} \binom{r}{3} \]
- For \( r = 3 \): \( \binom{3}{3} \binom{3}{3} = 1 \cdot 1 = 1 \)
- Total for \( s = 3 \): \( 1 \)

3. **Combine the Results**:
\[ f(3) = (12) + (6) + (1) = 19 \]

### Final Answer
Thus, the value of \( f(3) \) is \( \boxed{19} \).