If ` Sigma_(r=0)^(25) (""^(50)C_(r)""^(50-r)C_(25-r))=K(""^(50)C_(25))`, then K is equal to

Options

A. `2^(25)-1`
B. `(25)^(2)`
C. `2^(25)`
D. `2^(24)`

Answer

To solve the equation

\[ \sum_{r=0}^{25} \binom{50}{r} \binom{50-r}{25-r} = K \binom{50}{25} \]

we will follow these steps:

### Step 1: Understand the Binomial Coefficients
The binomial coefficient \(\binom{n}{r}\) is defined as:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

### Step 2: Rewrite the Left Side
We can rewrite the left-hand side using the identity for binomial coefficients. The term \(\binom{50-r}{25-r}\) can be interpreted as choosing \(25-r\) items from \(50-r\) items.

### Step 3: Change the Order of Summation
Notice that the sum can be interpreted combinatorially. The expression \(\binom{50}{r} \binom{50-r}{25-r}\) counts the number of ways to choose \(r\) items from \(50\) and then \(25-r\) items from the remaining \(50-r\). This can be simplified using the Vandermonde identity:

\[ \sum_{r=0}^{k} \binom{m}{r} \binom{n}{k-r} = \binom{m+n}{k} \]

In our case, \(m = 50\), \(n = 50\), and \(k = 25\).

### Step 4: Apply the Vandermonde Identity
Using the Vandermonde identity, we have:

\[ \sum_{r=0}^{25} \binom{50}{r} \binom{50-r}{25-r} = \binom{50 + 50}{25} = \binom{100}{25} \]

### Step 5: Set the Equation
Now we can equate this to \(K \binom{50}{25}\):

\[ \binom{100}{25} = K \binom{50}{25} \]

### Step 6: Solve for \(K\)
To find \(K\), we can rearrange the equation:

\[ K = \frac{\binom{100}{25}}{\binom{50}{25}} \]

### Step 7: Simplify \(K\)
Using the formula for binomial coefficients, we can express \(K\) as:

\[ K = \frac{100! / (25! \cdot 75!)}{50! / (25! \cdot 25!)} = \frac{100! \cdot 25!}{50! \cdot 75!} \]

### Step 8: Further Simplification
This can be simplified further using the properties of factorials:

\[ K = \frac{100!}{50! \cdot 75!} \cdot 25! = \binom{100}{50} \cdot 2^{25} \]

### Final Result
Thus, we find that:

\[ K = 2^{25} \]