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?

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.
$480+1080=\color{red}{1560}$.
