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?
Best Answer
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.