Number of ways to award 3 identical prizes to 20 students 
In how many ways can 3 identical prizes be awarded to 20 students, if (i) no student may win more than 1 prize. (ii) no student may win all 3 prizes.

For (i) I used the combination theory as three prizes are identical and order doesn’t matter, so my answer is $\binom{20}3$.
But for (ii), I tried to use equation of $20^3 - 20 $ but I thought that there would be some problems with the order since three prizes are identical, so I don’t know what to continue.
 A: For part (i), your answer of ${\large{\binom{20}{3}}}$ is correct.

For part (ii), since no student can win all $3$ prizes, it follows that there must be at least two students who each win at least one prize.

If there are $3$ such students, the number of qualifying distributions is  ${\large{\binom{20}{3}}}$, as in part (i).

If there are only $2$ such students, one of them must win two prizes, and the other student just one prize, hence the number of qualifying distributions is  $20{\,\cdot\,}19$,$\;$since there are $20$ choices for the student who wins two prizes, followed by $19$ choices for the student who wins just one prize.

Hence for part (ii), the total number of qualifying distributions is$\;{\large{\binom{20}{3}}}+(20{\,\cdot\,}19)$.
A: Your answer for part (i) is correct.
For part (ii), using stars and bars Th. 2, there are $\binom{3+20-1}{20-1} = 1540$ unrestricted distributions, but $20$ of hem are barred as no student can get all $3$ prizes, hence final answer $=\boxed{1520}$
Note
If you haven't yet learnt "stars and bars", you must make it a point to learn it immediately. It will prove extremely useful as you tackle more difficult problems of this type.
