If the number of ways in which 7 letters can be put in 7 envelopes such that exactly four letters are in wrong envelopes is N, then

Options

A. Number of divisors of N is 12
B. Sum of proper divisors of N is 624
C. Number of divisors of N which are divisible by 3 but not by 9 is 4
D. Sum of proper divisors of N is 309

Answer

To solve the problem of finding the number of ways in which 7 letters can be put in 7 envelopes such that exactly four letters are in wrong envelopes, we can follow these steps:

### Step 1: Choose the letters that will be in the correct envelopes
We need to select 3 letters out of the 7 that will be placed in their correct envelopes. The number of ways to choose 3 letters from 7 is given by the combination formula:

\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 2: Arrange the remaining letters in the wrong envelopes
Now, we have 4 letters that need to be placed in 4 envelopes, but they must all be placed incorrectly. This is known as a derangement. The formula for the number of derangements \( !n \) of \( n \) items is:

\[ !n = n! \left( \sum_{i=0}^{n} \frac{(-1)^i}{i!} \right) \]

For \( n = 4 \):

\[ !4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) \]

Calculating \( 4! \):

\[ 4! = 24 \]

Now calculating the sum:

\[ \frac{1}{0!} = 1, \quad \frac{1}{1!} = 1, \quad \frac{1}{2!} = 0.5, \quad \frac{1}{3!} \approx 0.1667, \quad \frac{1}{4!} = 0.0417 \]

So,

\[ 1 - 1 + 0.5 - 0.1667 + 0.0417 = 0.375 \]

Now, substituting back into the derangement formula:

\[ !4 = 24 \times 0.375 = 9 \]

### Step 3: Combine the results
Now, we multiply the number of ways to choose the 3 letters with the number of derangements of the remaining 4 letters:

\[ N = \binom{7}{3} \times !4 = 35 \times 9 = 315 \]

### Final Answer
The number of ways in which 7 letters can be put in 7 envelopes such that exactly four letters are in wrong envelopes is \( N = 315 \).

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