Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed? 
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$
Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$, $2$, or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn.
If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.
In the other case, two students receive $2$ awards and one student recieves $1$ award . We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{\textbf{(B) }150}$.