If `n > 3,` then `x y C_0-(x-1)(y-1)C_1+(x-2)(y-2)C_2-(x-3)(y-3)C_3+..............+(-1)^n(x-n)(y-n)C_n,` equals

Options

A. xyz
B. nxyz
C. #REF!
D. 0

Answer

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

\[ x y C_0 - (x-1)(y-1) C_1 + (x-2)(y-2) C_2 - (x-3)(y-3) C_3 + \ldots + (-1)^n (x-n)(y-n) C_n \]

where \( C_k \) represents the binomial coefficient \( \binom{n}{k} \).

### Step 1: Rewrite the expression

The expression can be rewritten as:

\[ \sum_{k=0}^{n} (-1)^k (x-k)(y-k) C_k \]

### Step 2: Expand the terms

We can expand \( (x-k)(y-k) \):

\[ (x-k)(y-k) = xy - (x+y)k + k^2 \]

### Step 3: Substitute the expansion back into the sum

Substituting this back into the summation gives:

\[ \sum_{k=0}^{n} (-1)^k \left( xy - (x+y)k + k^2 \right) C_k \]

This can be separated into three sums:

\[ xy \sum_{k=0}^{n} (-1)^k C_k - (x+y) \sum_{k=0}^{n} (-1)^k k C_k + \sum_{k=0}^{n} (-1)^k k^2 C_k \]

### Step 4: Evaluate each sum

1. **First sum**:
\[ \sum_{k=0}^{n} (-1)^k C_k = (1-1)^n = 0 \quad \text{(for } n > 0\text{)} \]

2. **Second sum**:
\[ \sum_{k=0}^{n} (-1)^k k C_k = n(-1)^{n-1} \quad \text{(using the identity for } k C_k\text{)} \]
However, since \( \sum_{k=0}^{n} (-1)^k C_k = 0 \), this term will also contribute 0.

3. **Third sum**:
\[ \sum_{k=0}^{n} (-1)^k k^2 C_k = n(n-1)(-1)^{n-2} \quad \text{(using the identity for } k^2 C_k\text{)} \]
Again, since the first sum is 0, this term will also contribute 0.

### Step 5: Combine the results

Since all three sums evaluate to 0, we have:

\[ xy \cdot 0 - (x+y) \cdot 0 + 0 = 0 \]

### Conclusion

Thus, the entire expression simplifies to:

\[ \boxed{0} \]