The sum `sumsum_(0leilejle10) (""^(10)C_(j))(""^(j)C_(r-1))` is equal to

Options

A. `2^(10) - 1`
B. `2^(10)`
C. `3^(10) - 1`
D. `3^(10)`

Explanation

To solve the problem, we need to evaluate the double summation:

\[ \sum_{0 \leq r \leq j \leq 10} \binom{10}{j} \binom{j}{r-1} \]

### Step-by-Step Solution:

1. **Understanding the Summation**:
The expression involves a double summation where \( j \) ranges from 0 to 10, and for each \( j \), \( r \) ranges from 0 to \( j \). The binomial coefficients \(\binom{10}{j}\) and \(\binom{j}{r-1}\) represent the number of ways to choose \( j \) items from 10 and \( r-1 \) items from \( j \), respectively.

2. **Changing the Order of Summation**:
We can rewrite the double summation by changing the order of summation. Instead of summing over \( r \) first, we can sum over \( j \) first:
\[ \sum_{j=0}^{10} \binom{10}{j} \sum_{r=0}^{j} \binom{j}{r-1} \]

3. **Evaluating the Inner Summation**:
The inner summation \(\sum_{r=0}^{j} \binom{j}{r-1}\) can be simplified. Notice that when \( r = 0 \), \(\binom{j}{-1} = 0\). Therefore, we can start the summation from \( r = 1 \):
\[ \sum_{r=1}^{j} \binom{j}{r-1} = \sum_{k=0}^{j-1} \binom{j}{k} = 2^j \]
where we used the identity that the sum of the binomial coefficients equals \( 2^j \).

4. **Substituting Back**:
Now, we substitute back into the original summation:
\[ \sum_{j=0}^{10} \binom{10}{j} 2^j \]

5. **Using the Binomial Theorem**:
According to the Binomial Theorem, we know that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
Setting \( n = 10 \) and \( x = 2 \), we have:
\[ (1 + 2)^{10} = 3^{10} \]

6. **Final Result**:
Therefore, the double summation evaluates to:
\[ \sum_{j=0}^{10} \binom{10}{j} 2^j = 3^{10} \]

### Conclusion:
The value of the given summation is \( 3^{10} \).