Question
In how many ways can $4$ prizes be distributed among $5$ students, when
  1. No student gets more than one prize?
  2. A student may get any number of prizes?
  3. No student gets all the prizes?

Answer

  1. $4$ prizes be distributed among $5$ students so that no student gets more than one prize can be done in
$^5\text{p}_4= \frac{5!}{(5-4)!}=\frac{5!}{(1)!}= 5! \ \text{ways}.$
  1. The first prize can be given away in $5$ ways as it may be given to anyone of the $5$ students. The second prize can also be given away in $5$ ways, since if may be obtained by the student who has already received a prize. Similarly, third and fourth prize can be given away in $5$ ways.
Hence, the number of ways in which all the prize can be given away $= 5 \times 5 \times 5 \times 5 = 625$
  1. Since any of the $5$ students may get all the prizes. So, the number of ways in which a student gets all the $4$ prizes is $5.$
So, the number of ways in which a student does not get all the prizes $= 625 - 5 = 620$

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