There are six letters `L_(1),L_(2),L_(3),L_(4),L_(5),L_(6)` and their corresponding six envelops `E_(1),E_(2),E_(3),E_(4),E_(5),E_(6)`. Letters having odd value can be put into odd valued envelopes and even valued letters can be put into even valued envelopes, so that no letter goes into the right envelopes, then number of arrangements equals.

Options

A. 6
B. 9
C. 44
D. 4

Answer

To solve the problem, we need to determine the number of ways to arrange six letters \( L_1, L_2, L_3, L_4, L_5, L_6 \) into their corresponding envelopes \( E_1, E_2, E_3, E_4, E_5, E_6 \) under the constraints that:

1. Odd valued letters \( L_1, L_3, L_5 \) can only go into odd valued envelopes \( E_1, E_3, E_5 \).
2. Even valued letters \( L_2, L_4, L_6 \) can only go into even valued envelopes \( E_2, E_4, E_6 \).
3. No letter can go into its corresponding envelope.

### Step-by-Step Solution:

**Step 1: Identify the Odd and Even Letters and Envelopes**
- Odd letters: \( L_1, L_3, L_5 \)
- Odd envelopes: \( E_1, E_3, E_5 \)
- Even letters: \( L_2, L_4, L_6 \)
- Even envelopes: \( E_2, E_4, E_6 \)

**Step 2: Count the Derangements for Odd Letters**
We need to find the number of derangements (permutations where no element appears in its original position) of the odd letters into the odd envelopes.

The formula for the number of derangements \( !n \) of \( n \) items is given by:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]

For \( n = 3 \) (odd letters):
\[ !3 = 3! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) \]
Calculating this:
\[ = 6 \left( 1 - 1 + 0.5 - \frac{1}{6} \right) = 6 \left( 0.5 - \frac{1}{6} \right) = 6 \left( \frac{3}{6} - \frac{1}{6} \right) = 6 \left( \frac{2}{6} \right) = 2 \]

So, there are \( 2 \) derangements for the odd letters.

**Step 3: Count the Derangements for Even Letters**
Similarly, we calculate the derangements for the even letters \( L_2, L_4, L_6 \).

For \( n = 3 \) (even letters):
\[ !3 = 3! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left( 1 - 1 + 0.5 - \frac{1}{6} \right) = 2 \]

So, there are also \( 2 \) derangements for the even letters.

**Step 4: Total Arrangements**
Since the arrangements of odd letters and even letters are independent, we multiply the number of derangements of odd letters by the number of derangements of even letters:
\[ \text{Total arrangements} = !3 \times !3 = 2 \times 2 = 4 \]

### Final Answer:
The total number of arrangements such that no letter goes into the correct envelope is \( 4 \).