If there were only 10 teams competing how many ways could 1st, 2nd, and 3rd place be awarded?

Question

5th grade math

Answer 1

If this question is asked in 5th grade, its answer should be explained at that level.

There are 10 teams to choose for 1st place, having done that there are 9 teams that could be in 2nd place, and that done there remain 8 teams that could be in 3rd place.  The total number of choices is 10× 9× 8 = 720

Answer 2

Hi, Patience,

 

This type of problem deals with "permutations" and "combinations", each of which is dealt with a little differently. In this case, we want to know, with 10 teams competing, how many ways 10 teams can be picked for 1st, 2nd, and 3rd. Since the order that we pick these teams DOES matter - meaning that picking team A (1st), B (2nd), and C (3rd) is DIFFERENT from picking team B (1st), A (2nd), C (3rd) - this is a PERMUTATION problem.

 

    (By contrast, an example of a COMBINATION would be if we had 10 people, and we wanted to find how many different committees of 3 people could be made. In this case, picking persons A, B, and C would really be the same as picking B, A, and C - if we are only concerned with the grouping, then the order really doesn't matter.)

 

We have 10 teams, and we want to know how many PERMUTATIONS of 3 can be made. Permutations are a little easier to calculate than combinations. For a PERMUTATION, we want to look at how many positions we are selecting, and how many choices there are for each of those selections.

 

So, for 3 selections, we want to look at:

 

    (# choices for 1st) x (# choices for 2nd) x (# choices for 3rd)

 

If there were originally 10 teams to choose from, then there are 10 options for 1st place. After one of them is chosen for 1st, there are only 9 choices for 2nd, and then there are only 8 choices for 3rd... and so on.

 

    (10) x (9) x (8) = 720

 

There are 720 different ways to pick 1st, 2nd, and 3rd place from among 10 original teams.

 

I hope this helps!

 

Andy