Question

there are 6 horses in a race. In how many different ways can places one, two and three occur?

Answer 1

that would \(6*5*4\)

Answer 2

it would be

Answer 3

since order matter you need permutation you need to choose 3 out of 6 in order

Answer 4

ok that seems logical, how would you set up the permutation? 6C3?

Answer 5

6P3
that one you wrote is combs

Answer 6

OK so we would have 6*5*4 as the numerator, what would the denominator be?

Answer 7

no the whole thing after simplifying is 6x5x4
is supposed to be 6!/(6-3)!

Answer 8

Ok, I have it now. Thanks I appreciate it!

Answer 9

welcome!