Question

A person writes letters to 6 friends and addresses envelopes. In how many ways can the letters be placed so that at least two go to wrong envelopes?

• A. ${}^6C_4 \cdot D_2$
• B. $\sum_{r=3}^6 {}^6C_r \cdot D_r$
• C. $\sum_{r=2}^6 {}^6C_r \cdot D_r$
• D. None of these

Hint:

To count “at least” derangements, use sum over $\binom{n}{r}D_r$

Answer 1

The Correct Option is C

We want at least 2 letters in wrong envelopes. This is the inclusion of all arrangements where 2 or more letters are deranged.
The number of such arrangements is: \[ \sum_{r=2}^6 \binom{6}{r} D_r \] Where $D_r$ is the number of derangements of $r$ items.