In how many ways $6$ letters can be placed in $6$ envelopes such that at most $3$ letters are placed in wrong envelopes.

Number of ways to place $6$ letters in $6$ envelopes such that at most $3$ letters are placed in wrong envelopes $=0$ letter is wrong envelope and $6$ in correct $+1$letter in wrong envelope and $5$ in correct $+ 2$ letters in wrong envelopes and $4$ are in correct $+ 3$ letters in wrong envelopes and $3$ in correct

$=1+0$(not possible to place $1$ in wrong envelope) $+6_{C_4}\times 1+6_{C_3}3!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}\right]$ 

$$\begin{gathered} =1+\frac{6\times 5}{2}+\frac{6\times 5\times 4}{6}\left(\frac{3!}{2!}-\frac{3!}{3!}\right) \\ =1+15+20\times 2=56 \end{gathered}$$