The value of `sum_(0leiltjle5) sum(""^(5)C_(j))(""^(j)C_(i))` is equal to `"_____"`

Explanation

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

\[ \sum_{0 \leq i < j \leq 5} \binom{5}{j} \binom{j}{i} \]

### Step 1: Understanding the Summation
The expression involves a double summation where \(i\) ranges from \(0\) to \(j-1\) and \(j\) ranges from \(1\) to \(5\). This means we will be summing over all pairs \((i, j)\) such that \(0 \leq i < j \leq 5\).

### Step 2: Changing the Order of Summation
We can switch the order of summation. Instead of summing over \(i\) first, we can sum over \(j\) first. Thus, we can rewrite the expression as:

\[ \sum_{j=1}^{5} \binom{5}{j} \sum_{i=0}^{j-1} \binom{j}{i} \]

### Step 3: Evaluating the Inner Sum
The inner sum \(\sum_{i=0}^{j-1} \binom{j}{i}\) represents the sum of the binomial coefficients for \(j\) and can be simplified using the binomial theorem. Specifically, we know that:

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

However, since we are summing from \(0\) to \(j-1\), we have:

\[ \sum_{i=0}^{j-1} \binom{j}{i} = 2^j - \binom{j}{j} = 2^j - 1 \]

### Step 4: Substituting Back into the Expression
Now substituting back into our expression, we have:

\[ \sum_{j=1}^{5} \binom{5}{j} (2^j - 1) \]

This can be split into two separate sums:

\[ \sum_{j=1}^{5} \binom{5}{j} 2^j - \sum_{j=1}^{5} \binom{5}{j} \]

### Step 5: Evaluating Each Sum
1. **First Sum**: The first sum \(\sum_{j=1}^{5} \binom{5}{j} 2^j\) can be evaluated using the binomial theorem:

\[ \sum_{j=0}^{5} \binom{5}{j} 2^j = (1 + 2)^5 = 3^5 = 243 \]

Since we only want the sum from \(j=1\) to \(5\), we subtract the \(j=0\) term:

\[ 243 - \binom{5}{0} \cdot 2^0 = 243 - 1 = 242 \]

2. **Second Sum**: The second sum \(\sum_{j=1}^{5} \binom{5}{j}\) is simply:

\[ \sum_{j=0}^{5} \binom{5}{j} = 2^5 = 32 \]

Again, we subtract the \(j=0\) term:

\[ 32 - 1 = 31 \]

### Step 6: Final Calculation
Now we can combine both results:

\[ 242 - 31 = 211 \]

Thus, the value of the original expression is:

\[ \boxed{211} \]