How many ways can 3 identical prizes be awarded to 98 potential winners?

[theRukus]

Homework Statement

How many was can 3 identical prizes be awarded to 98 potential winners?

Homework Equations

The Attempt at a Solution

Well. I know that if the prizes were unique, the first prize would have 98 possible winners, the second prize would have 97 possible winners, and the third prize would have 96 possible winners, totaling (98)(97)(96) possibilities. Since they are all identical, some number of solutions will be eliminated. Can anyone hint at how I should go about this?

 

[myth_kill]

can one winner have more that one prize?

 

[theRukus]

No, they're restricted to one prize each.

 

[HallsofIvy]

Yes, you are right that there are 98 ways to give the first prize to someone, 97 ways to give the second prize, and 96 ways to give the third. So if there were three different prizes, there would be 98(97)(96) ways to give out the three prizes.
(Which is equal, by the way, to
\frac{98!}{95!}= \frac{98!}{(98- 3)!}

But because the three prized are the same, it does not matter in which order we pick the three people. How many different orders are there for 3 people? So how many times is each set of 3 people being counted? Divide by that.

Does that answer look familiar? Here's another way to do the same problem: Label each person who is to be given a prize with an "A", label each person who is not with a "B". Different choices of people can be thought of as different orders for the "A"s and "B"s. How many different ways can you order 3 "A"s and 95 "B"s?