Distributing 4 distinct balls between 3 people 
In how many ways can you distribute 4 distinct balls between 3 people such that none of them gets exactly 2 balls?

This is what I did (by the inclusion–exclusion principle) and I'm not sure, would appreciate your feedback:
$$3^4-\binom{3}{1}\binom{4}{2}\binom{2}{1}^2+\binom{3}{2}\binom{4}{2}\binom{2}{1}$$

Method 1:  If no person receives exactly two balls, there are two possible cases.  One person receives all four balls or one person receives three balls and another person receives one ball.
Case 1:  One person receives all four balls.
There are three ways of selecting the person who receives all the balls.
Case 2:  There are three ways of selecting the person who receives three balls, $\binom{4}{3}$ ways of selecting which three of the four balls that person receives, and two ways of choosing the person who receives the remaining ball.  Consequently, there are 
$$\binom{3}{1}\binom{4}{3}\binom{2}{1}$$ 
of distributing three of the balls to one person and one ball to another person.
Note that Parcly Taxel and Jon Mark Perry obtained the factor of $$3! = \binom{3}{1}\binom{2}{1}$$ in this case by making the observation that there are $3!$ ways of distributing different numbers of balls to three different people.
Total:  Since these cases are disjoint, the number of ways of distributing four distinct balls to three people so that no person receives exactly two balls is 
$$\binom{3}{1} + \binom{3}{1}\binom{4}{3}\binom{2}{1}$$
Method 2:  Exclude those cases in which one person receives exactly two balls from the total.
Since both the people and the balls are distinct, there would be $3^4$ ways of distributing the balls to the people if there were no restrictions.  
There are two cases in which a person receives exactly two balls.  Either one person receives two balls, while the others receive one each, or two people receive two balls each.
Case 1:   One person receives two balls, while the others receive one each.  
There are three ways to select the person who receives two balls, $\binom{4}{2}$ ways of selecting the balls that person receives, two ways of selecting a ball for the remaining person whose name appears first alphabetically, and one way to give the remaining person the remaining ball.
$$\binom{3}{1}\binom{4}{2}\binom{2}{1}\binom{1}{1}$$
Case 2:  Two people each receive two balls.
There are $\binom{3}{2}$ ways to select which two people receive two balls, $\binom{4}{2}$ ways to choose which two balls are given to the selected person whose name appears first alphabetically, and one way to give the remaining two balls to the other selected person.
$$\binom{3}{2}\binom{4}{2}\binom{2}{2}$$
Total:  Since the cases are disjoint, the number of ways of distributing four distinct balls to three people so that no person receives exactly two balls is 
$$3^4 - \binom{3}{1}\binom{4}{2}\binom{2}{1} - \binom{3}{2}\binom{4}{2}$$
Method 3:  We use the Inclusion-Exclusion Principle.
There are $3^4$ ways of distributing four distinct balls to three people.
We must exclude those cases in which a person receives exactly two balls.
There are three ways of selecting a person to receive exactly two balls, $\binom{4}{2}$ ways of choosing which balls that person receives, and $2^2$ ways of distributing the remaining two balls to the other people.  Hence, there are 
$$\binom{3}{1}\binom{4}{2}2^2$$
ways of distributing the balls so that a person receives exactly two of them.  
However, we have counted those distributions in which two people receive two balls each twice, once for each of the people we designated as the person who receives two balls.  In Method 2, we calculated that the number of distributions in which two people each receive two balls is 
$$\binom{3}{2}\binom{4}{2}$$
Hence, by the Inclusion-Exclusion Principle, the number of ways of distributing four distinct balls to three people so that no person receives exactly two balls is 
$$3^4 - \binom{3}{1}\binom{4}{2}2^2 + \binom{3}{2}\binom{4}{2}$$ 
You calculated the last term incorrectly.
