How to distribute $k$ prizes in $p$ student with each student would receive $(k-1)$ prizes at most? 
A teacher decided to encourage the kids by distributing prizes to
  them. Each of the prizes was different from the other. The total
  number of prize was $k$, and total number of kids was $p$. To
  encourage the kids, the number of prizes was more than the number of
  kids. But, the teacher imposed a restriction on herself that each kid
  would receive $(k – 1)$ prizes at the most. How many ways she could
  distribute the prizes?

Answer is $p^k-p$
If I understood the problem correctly, it is no-where stated that a student can get no prize at all, and the answer seems to be using this assumption,however I am a bit confused why the answer is not $p^{k-1}$? 
If the restriction is that no student can get more than $(k-1)$ prizes then what is wrong with starting with $(k-1)$ distinct prizes and  distributing those into $p$ distinct groups (students)?
 A: Suppose there were no restrictions whatsoever and the problem simply said "How can you distribute $k$ prizes among $p$ students?".
For each prize, there are $p$ possibilities for the recipient of that prize. In total, that means there are $p \cdot p \cdot \dots \cdot p$ (with $k$ copies of $p$ in the product) possible ways to assign the prizes. This is the $p^k$ part.
Now let's impose the restriction that no student receives all the prizes (this is the same as saying no student receives more than $k-1$ prizes). Of the $p^k$ configurations we just enumerated, which ones are illegal under this new restriction? Exactly $p$ of them, since there is one illegal configuration for each student (namely, giving that student all the prizes).
Subtracting the illegal configurations from our previous enumeration, we get $p^k - p$ legal configurations.
A: The answer can easily be seen with this reasoning: there are p students and k prizes, so the total number of ways the prizes can be distributed is $p^k$ (for each of $k$ prizes, there are $p$ different students to whom it can be given). However, with the added constraint that no student can receive every prize, we must eliminate $p$ of those possibilities (one for when each student receives all of the prizes). Therefore the total number of ways is $p^k - p$.
We can't simply start with $k - 1$ prizes and distribute them to $p$ students because we would be missing the different cases that come from which student gets the last ($k^{th}$) prize.
