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 × 5 × 5 × 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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