Statement-1: the highest power of 3 in `.^(50)C_(10)` is 4.
Statement-2: If p is any prime number, then power of p in n! is equal to `[n/p]+[n/p^(2)]+[n/p^(3)]`+ . . ., where `[*]`
denotes the greatest integer function.

Options

A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1
C. Statement-1 is true, statement-2 is false
D. Statement-1 is false, statement-2 is true

Explanation

To solve the problem, we need to determine the highest power of 3 in the binomial coefficient \( \binom{50}{10} \) and verify the statements provided.

### Step-by-Step Solution

1. **Understanding the Binomial Coefficient**:
The binomial coefficient \( \binom{n}{r} \) is defined as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
For our case, we have:
\[ \binom{50}{10} = \frac{50!}{10! \cdot 40!} \]

2. **Finding the Highest Power of 3 in Factorials**:
We will use the formula for finding the highest power of a prime \( p \) in \( n! \):
\[ \text{Power of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
Here, \( p = 3 \).

3. **Calculating the Power of 3 in \( 50! \)**:
\[ \text{Power of 3 in } 50! = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor \]
- \( \left\lfloor \frac{50}{3} \right\rfloor = 16 \)
- \( \left\lfloor \frac{50}{9} \right\rfloor = 5 \)
- \( \left\lfloor \frac{50}{27} \right\rfloor = 1 \)
- \( \left\lfloor \frac{50}{81} \right\rfloor = 0 \)

Therefore, the total power of 3 in \( 50! \) is:
\[ 16 + 5 + 1 + 0 = 22 \]

4. **Calculating the Power of 3 in \( 40! \)**:
\[ \text{Power of 3 in } 40! = \left\lfloor \frac{40}{3} \right\rfloor + \left\lfloor \frac{40}{9} \right\rfloor + \left\lfloor \frac{40}{27} \right\rfloor + \left\lfloor \frac{40}{81} \right\rfloor \]
- \( \left\lfloor \frac{40}{3} \right\rfloor = 13 \)
- \( \left\lfloor \frac{40}{9} \right\rfloor = 4 \)
- \( \left\lfloor \frac{40}{27} \right\rfloor = 1 \)
- \( \left\lfloor \frac{40}{81} \right\rfloor = 0 \)

Therefore, the total power of 3 in \( 40! \) is:
\[ 13 + 4 + 1 + 0 = 18 \]

5. **Calculating the Power of 3 in \( 10! \)**:
\[ \text{Power of 3 in } 10! = \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor \]
- \( \left\lfloor \frac{10}{3} \right\rfloor = 3 \)
- \( \left\lfloor \frac{10}{9} \right\rfloor = 1 \)

Therefore, the total power of 3 in \( 10! \) is:
\[ 3 + 1 = 4 \]

6. **Finding the Power of 3 in \( \binom{50}{10} \)**:
Now we can find the power of 3 in \( \binom{50}{10} \):
\[ \text{Power of 3 in } \binom{50}{10} = \text{Power of 3 in } 50! - \text{Power of 3 in } 40! - \text{Power of 3 in } 10! \]
\[ = 22 - 18 - 4 = 0 \]

7. **Conclusion**:
- Statement 1 claims the highest power of 3 in \( \binom{50}{10} \) is 4, which is **false**.
- Statement 2 is true as it correctly describes the formula for finding the power of a prime in factorials.

### Final Answer:
- Statement 1 is false.
- Statement 2 is true.