The value of r for which
`.^(20)C_(r ), .^(20)C_(r - 1) .^(20)C_(1) + .^(20)C_(2) + …… + .^(20)C_(0) .^(20)C_(r )` is maximum, is

Options

A. 20
B. 15
C. 11
D. 10

Answer

To solve the problem, we need to find the value of \( r \) for which the expression

\[ \binom{20}{r} \cdot \binom{20}{r-1} \cdot \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{0} \cdot \binom{20}{r} \]

is maximized.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be interpreted as a sum of products of binomial coefficients. The first part, \( \binom{20}{r} \cdot \binom{20}{r-1} \cdot \binom{20}{1} \), represents a specific combination of selections from a set of 20 items.

2. **Using Binomial Theorem**:
We can use the Binomial Theorem to expand \( (1 + x)^{20} \):
\[ (1 + x)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^k \]
This gives us the coefficients of \( x^k \) as \( \binom{20}{k} \).

3. **Identifying the Maximum**:
The maximum value of \( \binom{20}{k} \) occurs at \( k = 10 \) (since \( n = 20 \) is even). This is due to the symmetric property of binomial coefficients.

4. **Finding the Value of \( r \)**:
To maximize the expression, we need to find \( r \) such that both \( \binom{20}{r} \) and \( \binom{20}{r-1} \) are maximized. Since \( \binom{20}{k} \) is maximized at \( k = 10 \), we can set \( r = 10 \) or \( r = 9 \) (since \( r-1 \) would then be 9 or 10).

5. **Conclusion**:
Thus, the value of \( r \) that maximizes the expression is \( r = 10 \).

### Final Answer:
The value of \( r \) for which the expression is maximum is \( r = 10 \).