In how many ways can $6$ prizes be distributed among $4$ persons such that each one gets at least one prize? My understanding:
First select $4$ prizes and distribute it among the $4$ people in $^6C_4\times4!$ ways and then distribute the remaining $2$ prizes in two cases: when $2$ people have $2$ prizes: $\frac{4!}{2!}$ways or when one person has $3$ prizes: $4$ ways. So totally: $^6C_4\times4!\times( \frac{4!}{2!}+4)$ ways $= 5760$ ways.
However the answer is $1560$ ways.
How?
 A: If the prizes are all different....
There are $4^6$ ways to distribute 6 prizes to 4 people without restrictions.
Rather than making sure everyone gets one, count the ways that no one gets one and subtract.  But you have to be careful about not double counting.
Inclusion-Exclusion:
$4^6 - {4\choose 3} 3^6 + {4\choose 2} 2^6 - {4\choose 1} 1^6$
A: We can also count this using a generating function approach.
Since the prizes are not identical, we compute the exponential generating function. To remove the possibility of anyone getting no prizes we use $e^x-1$ in each factor. The number of ways to distribute $6$ prizes among $4$ people is
$$
\begin{align}
6!\left[x^6\right]\left(e^x-1\right)^4
&=6!\left[x^6\right]\left(\binom44e^{4x}-\binom43e^{3x}+\binom42e^{2x}-\binom41e^{x}+\binom{4}{0}\right)\\
&=6!\left(\binom44\frac{4^6}{6!}-\binom43\frac{3^6}{6!}+\binom42\frac{2^6}{6!}-\binom41\frac{1^6}{6!}+\binom40\frac{0^6}{6!}\right)\\
&=\binom444^6-\binom433^6+\binom422^6-\binom411^6+\binom400^6\\[6pt]
&=1560
\end{align}
$$
A: There are two possibilities:  either one person receives three prizes and each of the other people receives one prize or two people each receive two prizes and the other two people each receive one prize.
One person receives three prizes and each of the other people receives one prize:  Select which of the four people receives three prizes.  Select which three of the six prizes that person receives.  Distribute the remaining three prizes to the remaining three people so that each person receives one prize.  There are
$$\binom{4}{1}\binom{6}{3}3!$$
such distributions.
Two people each receive two prizes and the other two people each receive one prize:  Select which two people receive two prizes each.  Select two of the six prizes for the younger of the two people who receive two prizes each.  Select two of the remaining four prizes for the older of the two people who receive two prizes each.  Distribute the remaining two prizes to the remaining two people so that each person receives one prize each.  There are
$$\binom{4}{2}\binom{6}{2}\binom{4}{2}2!$$
such distributions.
Since the two cases are mutually exclusive and exhaustive, there are a total of
$$\binom{4}{1}\binom{6}{3}3! + \binom{4}{2}\binom{6}{2}\binom{4}{2}2!$$
ways to distribute six prizes to four people so that each person receives at least one prize.
What errors did you make in your calculations?
You selected four of the six prizes to distribute to the four people, so you left out a factor of $\binom{6}{4}$.  This would make your answer too large.  The reason your answer would be too large is that you are counting each distribution in which one person receives three prizes three times, once for each way of designating one of those prizes, as the prize that person receives with the other two prizes as additional prizes.  Similarly, you count each distribution in which two people each receive two prizes four times, once for each way of designating one of the prizes each person receives as the the prize that person receives and the other prize that person receives as the additional prize. Notice that 
$$\color{blue}{\binom{6}{4}}4!\left[\frac{4!}{2!} + 4\right] = \color{red}{\binom{2}{1}^2}\binom{4}{2}\binom{6}{2}\binom{4}{2}2! + \color{red}{\binom{3}{1}}\binom{4}{1}\binom{6}{3}3!$$
A: Someone could get $3$ prizes & everyone else gets $1$: There are $4$ ways to choose the person who gets $3$ prizes and then $   (6 \times  5\times 4)/3! \times 3 \times 2 \times 1$ ways to distribute the prizes. So $480$ ways in this case.
OR 
Two people get $2$ prizes each & everyone else gets $1$: There are $6$ ways to choose the people who get $2$ prizes and then $  (6 \times  5)/2! \times (4 \times 3 )/2! \times 2 \times 1$ ways to distribute the prizes. So $1080$ ways in this case.
$480+1080=\color{red}{1560}$.
