factorial.. with replacement?

[masspt]

Here is the problem.. (literally..)
In your stats class, the teacher has three drawing for prizes (a, b , c). There are 15 students in class

a. if same person can win more than once, in how many different ways
can you choose the 3 names from the 15 students.

I am assuming this is WITH replacement...

without replacement you would use 15!/ (15-3)! * (3!) or 455
how do you factor back in the names already picked.. or am I way off
track?

b. If the selections above are random, what is the probability that the same
student wins all 3 prizes..

is this right (1/15)(1/15)(1/15) = 1/3375 ??

Thanks for your time.. M

[JohnM]

a. since you can replace the name after each draw, there are 15*15*15 ways that the names can be drawn

b. 15 divided by the answer in a. (since any of the 15 students can win all 3)
--> in other words, if you wrote out all the combinations from a., you would find that there are 15 instances of 1 student winning all 3 prizes

[masspt]

Thanks.. yet so obvious now..