Eight people are entered in a race. if there are no ties, in how many ways can the first three places come out? a.336 b.368 c.672 d.512 explain how to do this problem.

The first thing to do is to calculate how many ways you can choose 3 people from a set of eight.  In order to do this, we need to use the attached formula.
(The letter 'n' stands for the entire set and 'r' stands for the number of objects we wish to choose.)
So we wish to choose 3 people ('r') form a set of 8 ('n')
combinations = n! / r! * (n - r)!
combinations = 8 ! / (3! * 5!)
combinations = 8 * 7 * 6 * 5! / (3!) * (5!)
combinations = 8 * 7 * 6 / 3 * 2
combinations = 56

Now of those 56 combinations, the 3 people can finish in 6 different ways.
For example, persons A, B and C could finish
ABC or ACB or BAC or BCA or CAB or CBA

So to get the TOTAL combinations we multiply 56 * 6 which equals
336 so the answer is (a)

It should be a or d i’m not sure