Number of persons that can be the first = 8
Number of persons that can be the second = 7 (because 1 person has finished first)
Number of persons that can be the third = ....
Have any ideas to continue?
Eight runners are entered in the 1000-meter run. How many different first, second, and third place finishes could possibly occur? I have no idea how to start to help my daughter with this problem. Thanks for any help you can give this math-challenged mom.
we found a combinations/permutations calculator on-line, so we said:
Types to Choose (meaning how many runners): 8 = n
Number Chosen (meaning first place, second place and third): 3 = r
and repetition is not allowed meaning we used this formula:
n!/(n-r)! = 8!/(8-3)!
Now of course we needed to find out that ! is a factorial
8! = 8x7x6x5x4x3x2x1 or 40,230
(8-3) = 5! = 5x4x3x2x1 or 120
40,230/120 = 336 possible combinations
Are we correct?