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

Explanation

To solve the problem \( \sum_{r=0}^{3} \binom{8}{r} \left( \binom{5}{r+1} - \binom{4}{r} \right) \), we will break it down step by step.

### Step 1: Rewrite the Summation
We start with the expression:
\[ \sum_{r=0}^{3} \binom{8}{r} \left( \binom{5}{r+1} - \binom{4}{r} \right) \]
This can be separated into two sums:
\[ \sum_{r=0}^{3} \binom{8}{r} \binom{5}{r+1} - \sum_{r=0}^{3} \binom{8}{r} \binom{4}{r} \]

### Step 2: Evaluate the First Sum
For the first sum, we can use the identity:
\[ \sum_{r=0}^{n} \binom{m}{r} \binom{n}{k-r} = \binom{m+n}{k} \]
Here, we need to adjust the indices. We can change the index of the first sum by letting \( s = r + 1 \), which gives \( r = s - 1 \). Thus, the limits change from \( r = 0 \) to \( r = 3 \) into \( s = 1 \) to \( s = 4 \):
\[ \sum_{s=1}^{4} \binom{8}{s-1} \binom{5}{s} \]
This can be rewritten as:
\[ \sum_{s=0}^{4} \binom{8}{s-1} \binom{5}{s} = \sum_{s=0}^{4} \binom{8}{s} \binom{5}{s} = \binom{13}{4} \]

### Step 3: Evaluate the Second Sum
For the second sum, we can directly use the identity:
\[ \sum_{r=0}^{n} \binom{m}{r} \binom{n}{k-r} = \binom{m+n}{k} \]
Thus,
\[ \sum_{r=0}^{3} \binom{8}{r} \binom{4}{r} = \binom{12}{3} \]

### Step 4: Combine the Results
Now we combine the results of both sums:
\[ \sum_{r=0}^{3} \binom{8}{r} \binom{5}{r+1} - \sum_{r=0}^{3} \binom{8}{r} \binom{4}{r} = \binom{13}{4} - \binom{12}{3} \]

### Step 5: Calculate the Binomial Coefficients
Calculating the binomial coefficients:
\[ \binom{13}{4} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715 \]
\[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

### Step 6: Final Result
Thus, the final result is:
\[ 715 - 220 = 495 \]

Therefore, the value of the summation is:
\[ \boxed{495} \]