If `f(m) = sum_(i=0)^m ({:(30),(30-i):})({:(20),(m-i):})` where `({:(p),(q):})= ""^(p)C_(q)`, then

Options

A. maximum value of `f(m)` is `.^(50)C_(25)`
B. `f(0) + f(1)+"….."+f(50) = 2^(50)`
C. `f(m)` is always divisible by `50(1 le m le 49)`
D. The value of `underset(m=0)overset(50)sum(f(m))^(2) = .^(100)C_(50)`

Explanation

To solve the problem, we start with the function defined as:

\[ f(m) = \sum_{i=0}^{m} \binom{30}{30-i} \binom{20}{m-i} \]

This can be rewritten using the property of binomial coefficients, \(\binom{n}{r} = \binom{n}{n-r}\):

\[ f(m) = \sum_{i=0}^{m} \binom{30}{i} \binom{20}{m-i} \]

### Step 1: Identify the Range of \(m\)

Since \(\binom{20}{m-i}\) is defined only when \(0 \leq m-i \leq 20\), we conclude that \(m\) must satisfy:

\[ m \leq 20 \]

### Step 2: Apply the Vandermonde Identity

The sum can be interpreted using the Vandermonde identity, which states:

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

In our case, we can apply this identity with \(m = 30\), \(n = 20\), and \(r = m\):

\[ f(m) = \sum_{i=0}^{m} \binom{30}{i} \binom{20}{m-i} = \binom{30 + 20}{m} = \binom{50}{m} \]

### Step 3: Conclusion about \(f(m)\)

Thus, we have:

\[ f(m) = \binom{50}{m} \]

### Step 4: Analyze the Options

Now, we need to analyze the options given in the problem statement.

1. **Option A**: The maximum value of \(f(m)\) is \(\binom{50}{25}\) (this is incorrect since the maximum occurs at \(m = 25\) but \(m\) can only go up to 20).

2. **Option B**: \(f(0) + f(1) + \ldots + f(50) = 2^{50}\) (this is incorrect since \(f(m)\) is defined only for \(m \leq 20\)).

3. **Option C**: \(f(m)\) is divisible by 50 for \(1 \leq m \leq 49\) (this is correct; since \(f(m) = \binom{50}{m}\), and for \(1 \leq m \leq 49\), \(\binom{50}{m}\) is divisible by 50).

4. **Option D**: \(\sum_{m=0}^{50} f(m)^2 = \binom{100}{50}\) (this is incorrect since \(f(m)\) is defined only for \(m \leq 20\)).

### Final Answer

The correct option is **C**.