If x, y and 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!y!)/(r!)`
B. `((x!y!))/(r!)`
C. `""^(x+y)C_(r)`
D. `""^(xy)C_(r)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \binom{x}{r} + \binom{x}{r-1} \cdot \binom{y}{1} + \binom{x}{r-2} \cdot \binom{y}{2} + \ldots + \binom{y}{r} \]

### Step 1: Understand the Expression
The expression consists of a sum of products of binomial coefficients. Each term in the sum involves choosing \( r \) items from \( x \) and \( k \) items from \( y \), where \( k \) varies from \( 0 \) to \( r \).

### Step 2: Rewrite the Expression
We can rewrite the expression in a more manageable form. We notice that the sum can be expressed as:

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

This means we are summing over \( k \) where \( k \) runs from \( 0 \) to \( r \).

### Step 3: Use the Binomial Theorem
We can use the binomial theorem, which states that:

\[ (1 + a)^n = \sum_{k=0}^{n} \binom{n}{k} a^k \]

We can apply this theorem to our expression by considering:

\[ (1 + a)^x \cdot (1 + b)^y \]

### Step 4: Expand the Expression
Expanding \( (1 + a)^x \) and \( (1 + b)^y \):

\[ (1 + a)^x = \sum_{i=0}^{x} \binom{x}{i} a^i \]
\[ (1 + b)^y = \sum_{j=0}^{y} \binom{y}{j} b^j \]

### Step 5: Combine the Two Expansions
Now, we can combine these two expansions:

\[ (1 + a)^x \cdot (1 + b)^y = \sum_{i=0}^{x} \sum_{j=0}^{y} \binom{x}{i} \binom{y}{j} a^i b^j \]

### Step 6: Find the Coefficient of \( a^r \)
We need to find the coefficient of \( a^r \) in this combined expansion. The coefficient of \( a^r \) is given by:

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

### Step 7: Result
By the binomial theorem, the coefficient of \( a^r \) in \( (1 + a + b)^{x+y} \) is:

\[ \binom{x+y}{r} \]

Thus, we conclude that:

\[ S = \binom{x+y}{r} \]

### Final Answer
The final result is:

\[ \binom{x+y}{r} \]