Number of ways three awards can be given to 5 students Three distinct awards are to be given to a group of five students. In how many ways can this be done if (a) no student receives more than one award and (b) no student receives more than two awards?
My attempt:
(a)$$\binom{5}{1}\binom{4}{1}\binom{3}{1}=5\cdot4\cdot3=60$$
(b) $$60+\binom{5}{2}\binom{3}{1}+\binom{5}{1}\binom{3}{2}=105$$
Is this correct? I am pretty sure that part (a) is correct, but not about part (b). Also, if (b) is correct, is there an easier way to solve it?
 A: a) The first award can go to one of $5$ students, the second to $4$ and the third to $3$, hence your answer of $60$ is correct.
b) For each award, choose a student it goes to. This results in $5\times5\times5=125$ ways. Remove the $5$ ways where all $3$ awards go to the same person, and it results in $120$ ways in total.
Another way by cases: There are $60$ ways the awards go to 3 students, if the awards go to different students. If a student gets 2 awards, there 3 ways to split the awards $2$ to $1$. The group of $2$ can be given to one in $5$ students, and the group of $1$ can be given to one in $4$ students, making a total of $60+3\times4\times5=120$ ways.
A: Your answer to the first question is correct.
While I would do the second question the way Element118 did it, we can also consider cases.
If no student can receive all three awards, there are two possibilities:


*

*Each award is given to a different student.  

*One of the five students receives two of the three awards and one of the other four students receives the third award.


You calculated the answer to the first case.  For the second case, there are five ways to select the student who receives two awards, $\binom{3}{2}$ ways to select which two of the three awards that student receives, four ways to select the student who receives the third award, and one way to give that student that award.  Hence, the number of ways in which one of the students can receive two of the three awards and another student receives the third award is 
$$\binom{5}{1}\binom{3}{2}\binom{4}{1}$$
Hence, the number of ways in which the awards can be distributed so that no student receives all three wards is $$\binom{5}{1}\binom{4}{1}\binom{3}{1} + \binom{5}{1}\binom{3}{2}\binom{4}{1}$$
