Question

A person writes letters to \(6\) friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?
\[ \text{Notation: } D_n=n!\left(\sum_{i=0}^{n}\frac{(-1)^i}{i!}\right) \]

• A. \({}^6C_4D_2\)
• B. \(\displaystyle\sum_{r=3}^{6}{}^6C_{6-r}\cdot D_r\)
• C. \(\displaystyle\sum_{r=2}^{6}{}^6C_{6-r}\cdot D_r\)
• D. \({}^6C_1D_5+{}^6C_0D_6\)

Hint:

When exactly \(r\) objects are wrongly placed, first choose those related fixed or correct positions, then arrange the wrong ones using derangements \(D_r\).

Answer 1

The Correct Option is C

Step 1: Understand the condition.
There are \(6\) letters and \(6\) corresponding envelopes.
The condition says that at least two letters must be placed in wrong envelopes.
This means the number of wrongly placed letters can be
\[ 2,3,4,5,\text{ or }6 \]

Step 2: Choose correctly placed letters first.
Suppose exactly \(r\) letters are wrongly placed.
Then the remaining \(6-r\) letters are correctly placed.
The number of ways to choose these correctly placed letters is
\[ {}^6C_{6-r} \]

Step 3: Derange the wrongly placed letters.
After choosing \(6-r\) correctly placed letters, the remaining \(r\) letters must all go into wrong envelopes among themselves.
The number of ways to do this is the derangement number
\[ D_r \]
Therefore, for exactly \(r\) wrongly placed letters, the number of arrangements is
\[ {}^6C_{6-r}D_r \]

Step 4: Add all possible cases.
Since at least two letters are wrongly placed,
\[ r=2,3,4,5,6 \]
Hence, the total number of ways is
\[ \sum_{r=2}^{6}{}^6C_{6-r}D_r \]

Step 5: Final conclusion.
Therefore, the required number of ways is
\[ \boxed{\sum_{r=2}^{6}{}^6C_{6-r}D_r} \]