If `"^(n)C_(0)-^(n)C_(1)+^(n)C_(2)-^(n)C_(3)+...+(-1)^(r )*^(n)C_(r )=28` , then `n` is equal to ……

Options

A. `7`
B. `8`
C. `9`
D. `11`

Explanation

To solve the equation

\[ ^{n}C_{0} - ^{n}C_{1} + ^{n}C_{2} - ^{n}C_{3} + \ldots + (-1)^{r} \cdot ^{n}C_{r} = 28, \]

we can use the concept of binomial coefficients and the binomial theorem.

### Step-by-Step Solution:

1. **Understanding the Expression**: The given expression can be rewritten using the binomial theorem. The sum can be expressed as:
\[ \sum_{k=0}^{r} (-1)^{k} \cdot ^{n}C_{k} = (1 - 1)^{n} + \text{(higher order terms)} \]

2. **Using Binomial Theorem**: The expression can be simplified using the binomial expansion:
\[ (1 - 1)^{n} = 0^{n} = 0 \quad \text{(for } n > 0\text{)} \]
However, we need to consider the terms up to \( r \):
\[ \sum_{k=0}^{r} (-1)^{k} \cdot ^{n}C_{k} = (-1)^{r} \cdot ^{n-1}C_{r} \quad \text{(if } r < n\text{)} \]

3. **Setting Up the Equation**: We can equate this to the given value:
\[ (-1)^{r} \cdot ^{n-1}C_{r} = 28 \]

4. **Considering the Sign**: Since the left side can be either positive or negative depending on \( r \), we can assume \( r \) is even (to keep the left side positive):
\[ ^{n-1}C_{r} = 28 \]

5. **Finding Combinations**: We need to find \( n \) such that:
\[ ^{n-1}C_{r} = 28 \]

6. **Finding Possible Values**: The value 28 can be expressed as:
\[ ^{8}C_{2} = \frac{8 \times 7}{2 \times 1} = 28 \]
This means \( n-1 = 8 \) and thus \( n = 9 \).

### Final Answer:
\[ n = 9 \]