The sum `sum_(m)^(i=0) ({:(10),(i):})({:(20),(m-i):})`, (where `({:(p),(q):})= 0`, if `p lt q`) is maximum when `'m'` is

Options

A. 5
B. 10
C. 15
D. 20

Answer

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

\[ \sum_{i=0}^{m} \binom{10}{i} \binom{20}{m-i} \]

is maximized. Here, \(\binom{p}{q} = 0\) if \(p < q\).

### Step-by-Step Solution:

1. **Understanding the Sum**: The expression \(\sum_{i=0}^{m} \binom{10}{i} \binom{20}{m-i}\) represents the sum of products of binomial coefficients. This can be interpreted combinatorially as choosing \(i\) items from a set of 10 and \(m-i\) items from a set of 20.

2. **Using the Binomial Theorem**: The sum can be rewritten using the binomial theorem. The term \(\binom{10}{i}\) corresponds to the expansion of \((1+x)^{10}\) and \(\binom{20}{m-i}\) corresponds to \((1+x)^{20}\). Therefore, we can express the sum as the coefficient of \(x^m\) in the expansion of:

\[ (1+x)^{10} \cdot (1+x)^{20} = (1+x)^{30} \]

3. **Finding the Coefficient**: We need to find the coefficient of \(x^m\) in \((1+x)^{30}\), which is given by \(\binom{30}{m}\).

4. **Maximizing the Coefficient**: The binomial coefficient \(\binom{30}{m}\) is maximized when \(m\) is as close to \(n/2\) as possible, where \(n\) is the total number of items (in this case, 30). Since \(30\) is even, the maximum occurs at:

\[ m = \frac{30}{2} = 15 \]

5. **Conclusion**: Thus, the value of \(m\) that maximizes the sum is:

\[ m = 15 \]