`sum_(r=0)^(4) (-1)^(r )""^(16)C_(r)` is divisible by :

Options

A. 5
B. 7
C. 11
D. 13

Answer

To solve the problem \( \sum_{r=0}^{4} (-1)^{r} \binom{16}{r} \), we will evaluate the expression step by step.

### Step 1: Write the summation explicitly
We start by writing out the summation:
\[ \sum_{r=0}^{4} (-1)^{r} \binom{16}{r} = (-1)^{0} \binom{16}{0} + (-1)^{1} \binom{16}{1} + (-1)^{2} \binom{16}{2} + (-1)^{3} \binom{16}{3} + (-1)^{4} \binom{16}{4} \]

### Step 2: Calculate each term
Now we calculate each term:
- For \( r = 0 \):
\[ (-1)^{0} \binom{16}{0} = 1 \cdot 1 = 1 \]
- For \( r = 1 \):
\[ (-1)^{1} \binom{16}{1} = -1 \cdot 16 = -16 \]
- For \( r = 2 \):
\[ (-1)^{2} \binom{16}{2} = 1 \cdot \frac{16 \times 15}{2} = 120 \]
- For \( r = 3 \):
\[ (-1)^{3} \binom{16}{3} = -1 \cdot \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = -560 \]
- For \( r = 4 \):
\[ (-1)^{4} \binom{16}{4} = 1 \cdot \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820 \]

### Step 3: Combine the results
Now we combine all the calculated terms:
\[ 1 - 16 + 120 - 560 + 1820 \]

Calculating this step by step:
1. \( 1 - 16 = -15 \)
2. \( -15 + 120 = 105 \)
3. \( 105 - 560 = -455 \)
4. \( -455 + 1820 = 1365 \)

Thus, the result of the summation is:
\[ \sum_{r=0}^{4} (-1)^{r} \binom{16}{r} = 1365 \]

### Step 4: Check divisibility
Now we need to check the divisibility of \( 1365 \):
- Dividing by \( 5 \): \( 1365 \div 5 = 273 \) (divisible)
- Dividing by \( 7 \): \( 1365 \div 7 = 195 \) (not divisible)
- Dividing by \( 11 \): \( 1365 \div 11 = 124.09 \) (not divisible)
- Dividing by \( 13 \): \( 1365 \div 13 = 105 \) (not divisible)

### Conclusion
The only number that \( 1365 \) is divisible by from the options given is \( 5 \).

Thus, the answer is:
\[ \text{The correct option is } 5. \]