IF x,y,r are positive integers then
`""^x C_r+""^x C_(r-1) . ""^ y C_1+ ""^x C_(r-2) ""^y C_2+…….+"^y C_r=`

Options

A. `(x!)/2`
B. `((x+y)!)/2`
C. `"^(x+y)C_r`
D. none of these

Answer

To solve the problem, we need to find the sum of the following expression:

\[ \sum_{k=0}^{r} \binom{x}{r-k} \binom{y}{k} \]

where \(x\), \(y\), and \(r\) are positive integers.

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficient \(\binom{n}{k}\) represents the number of ways to choose \(k\) elements from a set of \(n\) elements. We will use the properties of binomial coefficients to simplify the expression.

2. **Rewriting the Sum**:
We can rewrite the sum as:
\[ \sum_{k=0}^{r} \binom{x}{r-k} \binom{y}{k} \]

3. **Applying the Vandermonde Identity**:
The Vandermonde identity states that:
\[ \sum_{k=0}^{r} \binom{x}{r-k} \binom{y}{k} = \binom{x+y}{r} \]
This identity allows us to combine the binomial coefficients into a single coefficient.

4. **Final Expression**:
By applying the Vandermonde identity, we conclude that:
\[ \sum_{k=0}^{r} \binom{x}{r-k} \binom{y}{k} = \binom{x+y}{r} \]

5. **Conclusion**:
Therefore, the final result of the sum is:
\[ \binom{x+y}{r} \]