Probability

Quote:

Originally Posted by mauro21pl

Fife separate awards are to be presented to selected students from a class of 30 people. How many different outcomes are possible if:
a)a student can receive any number of awards
b)each student can receive at most 1 award
thanks

(A)
Well for each award, there are 30 possibilities for whom to give the award.
So there are $30^5$ ways of distributing five awards to 30 students.

(B)
If no student can receive more than a single award, then calculate it the following way:
The first award can go to any of the 30 students.
The second award can go to any of the 29 students who haven't yet received an award.
The third award can go to any of the 28 students who haven't yet received an award.
The fourth award can go to any of the 27 students who haven't yet received an award.
Finally, the fifth award can go to any of the 26 students who haven't yet received an award.

So there are $30 \cdot 29 \cdot 28 \cdot 27 \cdot 26$ ways of doing this.

If you are familiar with binomial notation, this can be written as $5! \binom{30}{5}$

EDIT: Are the awards all the same award, or are they individual? If they are all the same award and it doesn't matter who gets what, then divide the result in part (B) by $5!$ , which is the number of ways to permute the five award-winning students. However, from your wording "five separate awards..." it sounds like they aren't all the same award, in which case this last part is moot.