The value of `"^(12)C_(2)+^(13)C_(3)+^(14)C_(4)+...+^(999)C_(989)` is

Options

A. `"^(1000)C_(11)-12`
B. `"^(1000)C_(11)+12`
C. `"^(900)C_(11)-12`
D. `"^(1000)C_(989)`

Explanation

To solve the problem \( ^{12}C_{2} + ^{13}C_{3} + ^{14}C_{4} + \ldots + ^{999}C_{989} \), we can use the properties of binomial coefficients.

### Step-by-step Solution:

1. **Identify the pattern in the binomial coefficients**:
The given expression is a sum of binomial coefficients starting from \( ^{12}C_{2} \) to \( ^{999}C_{989} \). We can rewrite this in a more general form:
\[ \sum_{n=12}^{999} {n \choose n-10} \]
This is equivalent to:
\[ \sum_{n=12}^{999} {n \choose 2} \]

2. **Use the Hockey Stick Identity**:
According to the Hockey Stick Identity in combinatorics, we have:
\[ \sum_{k=r}^{n} {k \choose r} = {n+1 \choose r+1} \]
In our case, we can set \( r = 2 \) and \( n = 999 \):
\[ \sum_{k=2}^{999} {k \choose 2} = {1000 \choose 3} \]

3. **Adjust the limits of the summation**:
We need to adjust our summation to start from \( 12 \):
\[ \sum_{k=12}^{999} {k \choose 2} = {1000 \choose 3} - \sum_{k=2}^{11} {k \choose 2} \]

4. **Calculate the sum from \( k=2 \) to \( k=11 \)**:
We can calculate \( \sum_{k=2}^{11} {k \choose 2} \):
\[ {2 \choose 2} + {3 \choose 2} + {4 \choose 2} + \ldots + {11 \choose 2} \]
This is:
\[ 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 = 220 \]

5. **Calculate \( {1000 \choose 3} \)**:
Using the formula for combinations:
\[ {1000 \choose 3} = \frac{1000 \times 999 \times 998}{3 \times 2 \times 1} = 166167000 \]

6. **Combine the results**:
Now we can substitute back into our equation:
\[ \sum_{k=12}^{999} {k \choose 2} = {1000 \choose 3} - 220 = 166167000 - 220 = 166166780 \]

Thus, the final answer is:
\[ \boxed{166166780} \]