The sum `sum_(0 leq i)sum_(leq j leq 10) (10C_j)(jC_i)` is equal to

Options

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

Answer

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

\[ \sum_{i=0}^{j} \sum_{j=0}^{10} \binom{10}{j} \binom{j}{i} \]

### Step 1: Understand the summation

We have a double summation where \( j \) varies from 0 to 10, and for each \( j \), \( i \) varies from 0 to \( j \). The terms involved are binomial coefficients.

### Step 2: Change the order of summation

We can rewrite the double summation by changing the order of summation. This means we will first sum over \( j \) and then over \( i \):

\[ \sum_{j=0}^{10} \binom{10}{j} \sum_{i=0}^{j} \binom{j}{i} \]

### Step 3: Evaluate the inner summation

The inner summation \( \sum_{i=0}^{j} \binom{j}{i} \) represents the sum of the binomial coefficients for a fixed \( j \). According to the binomial theorem, this sum equals \( 2^j \):

\[ \sum_{i=0}^{j} \binom{j}{i} = 2^j \]

### Step 4: Substitute back into the outer summation

Now we substitute this result back into the outer summation:

\[ \sum_{j=0}^{10} \binom{10}{j} 2^j \]

### Step 5: Recognize the outer summation as a binomial expansion

The expression \( \sum_{j=0}^{10} \binom{10}{j} 2^j \) can be recognized as the expansion of \( (2 + 1)^{10} \) using the binomial theorem:

\[ \sum_{j=0}^{n} \binom{n}{j} x^j = (x + 1)^n \]

Here, \( n = 10 \) and \( x = 2 \):

\[ \sum_{j=0}^{10} \binom{10}{j} 2^j = (2 + 1)^{10} = 3^{10} \]

### Step 6: Final result

Thus, the value of the original double summation is:

\[ 3^{10} \]

### Conclusion

The final answer is \( 3^{10} \).

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