Envelope Question: Five letters addressed to individuals 1-5 are randomly placed in five addressed envelopes, one letter in each envelope. I'm trying to find the probability of:


*

*Exactly three letters go in the correct envelopes.

*Exactly two letters go in the correct envelopes

*No letters go in the correct envelopes


Here is my approach:
So there is clearly a total of 5! distinct ways of arranging the letters.


*

*If exactly three letters go in the correct envelopes, then there are $5 \choose 3$ ways of choosing the positions for the three correct envelopes, and for the remaining two letters, there are 2! ways of organizing them. Thus, probability = $\frac{{5 \choose 3} \cdot 2!}{5!}$.

*If exactly two letters go in the correct envelopes, then there are $5 \choose 2$ ways of choosing the positions for the two correct envelopes, and for the remaining three letters, there are 3! ways of organizing them. Thus, probability = $\frac{{5 \choose 2} \cdot 3!}{5!}$.

*I'm not really sure how to approach this problem. 
Any input would be great. 

There are not $2!$ ways to organize the lat two letters. There is only $1$ way. Because the second way of organizing them would be to put them in their correct envelopes, which wouldn't match up with the constraint of having exactly $3$ letters getting sent correctly.
A similar mistake was made in the second problem.
To find the number of ways no letters get put in the correct envelope, also known as finding the number of derangements, start by taking $5!$ and subtract off all the cases letter $1$ is in the correct spot, then letter $2$, letter $3$, etc. Then use the inclusion exclusion principle
