How many options are there to award gold, silver, and bronze medals to a group of $10$ athletes?

Is this permutation or combination, and is there repetition? I thought this would be a combination without repetition, where I found the formula to be: $\frac{n!}{(n-r)!} = \frac{10!}{7!} = 720$ I'm doubting that this is correct though. Did I do something wrong?

  • 3
    $\begingroup$ This is a permutation, and your answer is right. There are 10 choices for the gold, then 9 choices for the silver, and 8 possibilities for the bronze, giving $10\cdot9\cdot8=720$ $\endgroup$ – user84413 Feb 4 '16 at 17:15

This is fine as long as you care which athlete gets which medal (and I am sure they do).

  • $\begingroup$ why isn't it 3 possibilities for athlete one , 3 for two , ext, giving 3^10? since wording doesn't clarify only one of each medal are given $\endgroup$ – Randin Mar 7 '18 at 16:24

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