Math Question?

Question

Eight runners are entered in the 1000 meter run. How many different first, second and third place finishes could possibly occur. I need help finding the formula to use i.e.: P(E)..... and the way you do it. Please help.

I get what you are saying but how many first
second
third
place finishes can occur I have to have a number for each.

Answer 1

There are 8 choices for 1st, then 7 remaining who could come in 2nd, then 6 for third. The answer is 8*7*6, which is 336.

The formula is the number of permutations of 3 chosen from a set of 8:

P(8,3) =
8!/(8-3)! =
8*7*6*5*4*3*2*1 / (5*4*3*2*1) =
8*7*6

Answer 2

Any of 8 runners can win.
When one has won, that leaves 7 who can come second.
When the first two are over the line, that leaves 6 who can come third.
The total number of possibilities is:
8 * 7 * 6 = 336.
It is a permutation formula P(n,r) = n! / (n - r)!, with n = 8 and r = 3, as the sequence is improtant, but you don't really need to use a formula. You can work this out just using common sense.