If `n gt 3`, then `xyz^(n)C_(0)-(x-1)(y-1)(z-1)""^(n)C_(1)+(x-2)(y-2)(z-2)""^(n)C_(2)-`
`(x-3)(y-3)(z-3)""^(n)C_(3)+…..+(-1)^(n)(x-n)(y-n)(z-n)""^(n)C_(n)` equals :

Options

A. xyz
B. `x+y+z`
C. `xy+yz+zx`
D. 0

Answer

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

\[ xyz^{(n)}C_{0} - (x-1)(y-1)(z-1)^{(n)}C_{1} + (x-2)(y-2)(z-2)^{(n)}C_{2} - (x-3)(y-3)(z-3)^{(n)}C_{3} + \ldots + (-1)^{n}(x-n)(y-n)(z-n)^{(n)}C_{n} \]

### Step 1: Rewrite the expression
We can rewrite the expression in a more compact summation form:

\[ \sum_{r=0}^{n} (-1)^{r} (x-r)(y-r)(z-r) C_{r}^{(n)} \]

### Step 2: Expand the terms
We can expand the terms \((x-r)(y-r)(z-r)\):

\[ (x-r)(y-r)(z-r) = xyz - r(xy + xz + yz) + r^2(x+y+z) - r^3 \]

### Step 3: Substitute back into the summation
Substituting the expanded form back into the summation gives us:

\[ \sum_{r=0}^{n} (-1)^{r} \left( xyz - r(xy + xz + yz) + r^2(x+y+z) - r^3 \right) C_{r}^{(n)} \]

### Step 4: Separate the summation
We can separate the summation into four different parts:

1. \(xyz \sum_{r=0}^{n} (-1)^{r} C_{r}^{(n)}\)
2. \(- (xy + xz + yz) \sum_{r=0}^{n} (-1)^{r} r C_{r}^{(n)}\)
3. \(+ (x+y+z) \sum_{r=0}^{n} (-1)^{r} r^2 C_{r}^{(n)}\)
4. \(- \sum_{r=0}^{n} (-1)^{r} r^3 C_{r}^{(n)}\)

### Step 5: Evaluate each summation
1. **First Summation**: The first summation evaluates to 0 because it is the sum of binomial coefficients with alternating signs:

\[ \sum_{r=0}^{n} (-1)^{r} C_{r}^{(n)} = 0 \]

2. **Second Summation**: The second summation also evaluates to 0:

\[ \sum_{r=0}^{n} (-1)^{r} r C_{r}^{(n)} = 0 \]

3. **Third Summation**: The third summation evaluates to 0 as well:

\[ \sum_{r=0}^{n} (-1)^{r} r^2 C_{r}^{(n)} = 0 \]

4. **Fourth Summation**: The fourth summation evaluates to 0:

\[ \sum_{r=0}^{n} (-1)^{r} r^3 C_{r}^{(n)} = 0 \]

### Step 6: Combine the results
Since all four parts of the summation evaluate to 0, we conclude that the entire expression is equal to 0:

\[ xyz \cdot 0 - (xy + xz + yz) \cdot 0 + (x+y+z) \cdot 0 - 0 = 0 \]

### Final Result
Thus, the value of the given expression is:

\[ \boxed{0} \]