The value of `sum_(r=0)^(3) ""^(8)C_(r)(""^(5)C_(r+1)-""^(4)C_(r))` is `"_____"`.

Explanation

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

\[ \sum_{r=0}^{3} \binom{8}{r} \left( \binom{5}{r+1} - \binom{4}{r} \right) \]

### Step 1: Break down the summation

We can separate the summation into two parts:

\[ \sum_{r=0}^{3} \binom{8}{r} \binom{5}{r+1} - \sum_{r=0}^{3} \binom{8}{r} \binom{4}{r} \]

### Step 2: Evaluate the first summation

For the first summation, we can use the identity:

\[ \sum_{r=0}^{n} \binom{m}{r} \binom{n}{r+k} = \binom{m+n}{n+k} \]

In our case, we have \( m = 8 \), \( n = 5 \), and \( k = 1 \). Thus, we can write:

\[ \sum_{r=0}^{3} \binom{8}{r} \binom{5}{r+1} = \sum_{r=1}^{4} \binom{8}{r-1} \binom{5}{r} \]

This can be rewritten as:

\[ \sum_{s=0}^{3} \binom{8}{s} \binom{5}{s+1} = \binom{8+5}{5+1} = \binom{13}{6} \]

### Step 3: Evaluate the second summation

For the second summation, we can directly compute:

\[ \sum_{r=0}^{3} \binom{8}{r} \binom{4}{r} \]

This can also be evaluated using the Vandermonde identity:

\[ \sum_{r=0}^{n} \binom{m}{r} \binom{n}{k-r} = \binom{m+n}{k} \]

In our case, \( m = 8 \), \( n = 4 \), and \( k = 3 \):

\[ \sum_{r=0}^{3} \binom{8}{r} \binom{4}{r} = \binom{8+4}{3} = \binom{12}{3} \]

### Step 4: Calculate the values

Now we need to calculate the binomial coefficients:

1. Calculate \( \binom{13}{6} \):
\[ \binom{13}{6} = \frac{13!}{6! \cdot 7!} = 1716 \]

2. Calculate \( \binom{12}{3} \):
\[ \binom{12}{3} = \frac{12!}{3! \cdot 9!} = 220 \]

### Step 5: Combine the results

Now, substituting back into our separated summation:

\[ \sum_{r=0}^{3} \binom{8}{r} \left( \binom{5}{r+1} - \binom{4}{r} \right) = \binom{13}{6} - \binom{12}{3} = 1716 - 220 = 1496 \]

Thus, the final answer is:

\[ \text{The value is } 1496 \]