Statement -1 : Six different balls are put in three different boxes, no box being empty, the probability of putting balls in boxes in equal numbers is `1/6`
Statement-2 : Six letters are posted in 3 letter boxes. The probability that no letter box remains empty is `20/27`

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement -5
B. Statement-1 is true, Statement-2 is true, Statement-2 is Not a correct explanation for statement -5
C. Statement-1 is true, Statement-2 is False
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we will analyze both statements step by step.

### Statement 1:
**Six different balls are put in three different boxes, no box being empty, the probability of putting balls in boxes in equal numbers is `1/6`.**

1. **Total Ways to Distribute Balls:**
We need to find the total number of ways to distribute 6 different balls into 3 different boxes such that no box is empty. We can use the principle of inclusion-exclusion or directly calculate the distributions.

- The valid distributions of 6 balls into 3 boxes where no box is empty can be:
- (4, 1, 1)
- (3, 2, 1)
- (2, 2, 2)

2. **Calculating Each Distribution:**
- For (4, 1, 1):
- Choose 4 balls from 6: \( \binom{6}{4} = 15 \)
- The arrangement of (4, 1, 1) can be done in \( \frac{3!}{1!1!1!} = 3 \) ways.
- Total ways = \( 15 \times 3 = 45 \)

- For (3, 2, 1):
- Choose 3 balls from 6: \( \binom{6}{3} = 20 \)
- Choose 2 from remaining 3: \( \binom{3}{2} = 3 \)
- Arrangement of (3, 2, 1): \( \frac{3!}{1!1!1!} = 6 \)
- Total ways = \( 20 \times 3 \times 6 = 360 \)

- For (2, 2, 2):
- Choose 2 balls from 6: \( \binom{6}{2} = 15 \)
- Choose 2 from remaining 4: \( \binom{4}{2} = 6 \)
- Choose 2 from remaining 2: \( \binom{2}{2} = 1 \)
- Arrangement of (2, 2, 2): \( \frac{3!}{2!} = 3 \)
- Total ways = \( 15 \times 6 \times 1 \times 3 = 270 \)

3. **Total Valid Distributions:**
- Total ways = \( 45 + 360 + 270 = 675 \)

4. **Total Ways to Distribute 6 Balls into 3 Boxes:**
- Each ball can go into any of the 3 boxes: \( 3^6 = 729 \)

5. **Probability of Equal Distribution (2, 2, 2):**
- The number of ways to achieve the (2, 2, 2) distribution is 270.
- Probability = \( \frac{270}{729} = \frac{30}{81} = \frac{10}{27} \)

6. **Conclusion for Statement 1:**
- The probability of putting balls in boxes in equal numbers is not \( \frac{1}{6} \) but \( \frac{10}{27} \). Thus, Statement 1 is **false**.

### Statement 2:
**Six letters are posted in 3 letter boxes. The probability that no letter box remains empty is `20/27`.**

1. **Total Ways to Distribute Letters:**
- Similar to the first statement, we need to find the total number of ways to distribute 6 letters into 3 boxes such that no box is empty.

- Using the same distributions as in Statement 1:
- (4, 1, 1): 45 ways
- (3, 2, 1): 360 ways
- (2, 2, 2): 270 ways
- Total ways = \( 45 + 360 + 270 = 675 \)

2. **Total Ways to Distribute 6 Letters into 3 Boxes:**
- Total ways = \( 3^6 = 729 \)

3. **Probability that No Box is Empty:**
- Probability = \( \frac{675}{729} = \frac{25}{27} \)

4. **Conclusion for Statement 2:**
- The probability that no letter box remains empty is not \( \frac{20}{27} \) but \( \frac{25}{27} \). Thus, Statement 2 is **false**.

### Final Conclusion:
- Both statements are false based on the calculations. The correct probabilities are \( \frac{10}{27} \) for Statement 1 and \( \frac{25}{27} \) for Statement 2.