STATEMENT -1: The exponent of 7 in `""^(100)C_(50)` is 4.
STATEMENT -2 : The number of ways in which we can post 5 letters in 12 boxes is `12^(5)` .

Options

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

Answer

To solve the given problem, we need to verify two statements regarding combinations and arrangements.

### Step 1: Verify Statement 1
**Statement 1:** The exponent of 7 in \( \binom{100}{50} \) is 4.

To find the exponent of a prime \( p \) in \( n! \), we use the formula:
\[ \text{Exponent of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]

Here, \( n = 100 \) and \( p = 7 \).

1. **Calculate the exponent of 7 in \( 100! \):**
\[ \left\lfloor \frac{100}{7} \right\rfloor + \left\lfloor \frac{100}{49} \right\rfloor + \left\lfloor \frac{100}{343} \right\rfloor \]
- \( \left\lfloor \frac{100}{7} \right\rfloor = 14 \)
- \( \left\lfloor \frac{100}{49} \right\rfloor = 2 \)
- \( \left\lfloor \frac{100}{343} \right\rfloor = 0 \)

Thus, the total for \( 100! \) is:
\[ 14 + 2 + 0 = 16 \]

2. **Calculate the exponent of 7 in \( 50! \):**
\[ \left\lfloor \frac{50}{7} \right\rfloor + \left\lfloor \frac{50}{49} \right\rfloor \]
- \( \left\lfloor \frac{50}{7} \right\rfloor = 7 \)
- \( \left\lfloor \frac{50}{49} \right\rfloor = 1 \)

Thus, the total for \( 50! \) is:
\[ 7 + 1 = 8 \]

3. **Calculate the exponent of 7 in \( \binom{100}{50} \):**
\[ \text{Exponent of } 7 \text{ in } \binom{100}{50} = \text{Exponent of } 7 \text{ in } 100! - 2 \times \text{Exponent of } 7 \text{ in } 50! \]
\[ = 16 - 2 \times 8 = 16 - 16 = 0 \]

**Conclusion for Statement 1:** The exponent of 7 in \( \binom{100}{50} \) is 0, not 4. Therefore, Statement 1 is false.

### Step 2: Verify Statement 2
**Statement 2:** The number of ways in which we can post 5 letters in 12 boxes is \( 12^5 \).

1. Each letter can be placed in any of the 12 boxes.
2. The first letter has 12 choices, the second letter also has 12 choices, and this continues for all 5 letters.

Thus, the total number of ways is:
\[ 12 \times 12 \times 12 \times 12 \times 12 = 12^5 \]

**Conclusion for Statement 2:** The number of ways to post 5 letters in 12 boxes is indeed \( 12^5 \). Therefore, Statement 2 is true.

### Final Conclusion
- Statement 1 is false.
- Statement 2 is true.

The correct option is that Statement 1 is false and Statement 2 is true.

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