`sum_(r=0)^(18)""^(20)C_(r)""^(20)C_(r+2)=""^(n)C_R` , where `R gt 20` then `(n-R)/2` = ____

Answer

To solve the problem, we need to evaluate the summation

\[ \sum_{r=0}^{18} \binom{20}{r} \binom{20}{r+2} = \binom{n}{R} \]

where \( R > 20 \). We are tasked with finding \( \frac{n - R}{2} \).

### Step-by-step Solution:

1. **Understanding the Summation**:
The summation involves binomial coefficients. The term \( \binom{20}{r} \) represents the number of ways to choose \( r \) items from 20, and \( \binom{20}{r+2} \) represents the number of ways to choose \( r+2 \) items from 20.

2. **Using Binomial Theorem**:
We can use the binomial theorem to expand \( (1+x)^{20} \). The expansion gives us:

\[ (1+x)^{20} = \sum_{r=0}^{20} \binom{20}{r} x^r \]

Therefore, we can write:

\[ (1+x)^{20} \cdot (1+x)^{20} = (1+x)^{40} \]

3. **Finding the Coefficient**:
The coefficient of \( x^{20} \) in \( (1+x)^{40} \) can be found using the binomial coefficient:

\[ \binom{40}{20} \]

However, we need to consider the specific terms \( \binom{20}{r} \) and \( \binom{20}{r+2} \).

4. **Rewriting the Summation**:
The summation can be rewritten using the identity:

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

In our case, we need to adjust for the \( r+2 \) term. By changing the index of summation, we can express the original summation in terms of a single binomial coefficient.

5. **Calculating the Result**:
After manipulating the summation, we find that:

\[ \sum_{r=0}^{18} \binom{20}{r} \binom{20}{r+2} = \binom{40}{22} \]

This implies:

\[ \binom{n}{R} = \binom{40}{22} \]

6. **Identifying \( n \) and \( R \)**:
From the equation \( \binom{n}{R} = \binom{40}{22} \), we can set \( n = 40 \) and \( R = 22 \) since \( R > 20 \).

7. **Finding \( \frac{n - R}{2} \)**:
Now, we calculate:

\[ n - R = 40 - 22 = 18 \]

Therefore,

\[ \frac{n - R}{2} = \frac{18}{2} = 9 \]

### Final Answer:
\[ \frac{n - R}{2} = 9 \]