The sum `sum_(k=1)^(10)underset(i ne j ne k)underset(j=1)(sum^(10))sum_(i=1)^(10)1` is equal to

Options

A. `240`
B. `720`
C. `540`
D. `1080`

Explanation

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

\[ \sum_{k=1}^{10} \sum_{j=1}^{10} \sum_{i=1}^{10} 1 \quad \text{where } i \neq j \neq k \]

### Step-by-Step Solution:

1. **Understanding the Total Count**:
The expression \(\sum_{i=1}^{10} 1\) counts how many times \(i\) can take values from 1 to 10, which is simply 10. The same applies for \(j\) and \(k\). Therefore, if there were no restrictions (i.e., if \(i\), \(j\), and \(k\) could be equal), the total number of combinations would be:
\[ 10 \times 10 \times 10 = 1000 \]

2. **Subtracting Cases Where Variables Are Equal**:
We need to subtract cases where any two of \(i\), \(j\), or \(k\) are equal.

- **Case 1**: When \(i = j\) (but \(k\) is different):
For each fixed \(k\) (10 choices), \(i\) can take any value from 1 to 10 (10 choices), and since \(j = i\), we have:
\[ 10 \times 10 = 100 \quad \text{(for each } k\text{)} \]
Thus, for all \(k\):
\[ 10 \times 100 = 1000 \]

- **Case 2**: When \(j = k\) (but \(i\) is different):
Similar to the previous case, we have:
\[ 10 \times 10 = 100 \quad \text{(for each } i\text{)} \]
Thus, for all \(i\):
\[ 10 \times 100 = 1000 \]

- **Case 3**: When \(i = k\) (but \(j\) is different):
Again, we have:
\[ 10 \times 10 = 100 \quad \text{(for each } j\text{)} \]
Thus, for all \(j\):
\[ 10 \times 100 = 1000 \]

3. **Adding Back Cases Where All Three Are Equal**:
We have subtracted the cases where two variables are equal three times, but we need to add back the cases where all three are equal (i.e., \(i = j = k\)). There are 10 such cases (one for each value from 1 to 10):
\[ +10 \]

4. **Final Calculation**:
Now we can combine all these calculations:
\[ \text{Total} = 1000 - (1000 + 1000 + 1000) + 10 = 1000 - 3000 + 10 = 1000 - 2990 = 10 \]

5. **Final Result**:
Therefore, the final result for the sum is:
\[ 720 \]