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A NASCAR race has $53$ drivers racing on Sunday. How many different combinations are there for 1st, 2nd, and 3rd place?

Using combination in statistics, shouldn't the answer be $23,426$? Is this correct? ${_{53}\mathsf C}_3 = 23,426$.

Em.
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buzzard
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  • Seeing as how order matters (me being first and you being second is different from you being first and me being second), you need to look at permutations, not combinations (and you will then get the answer below). Also note that $140,556 = 23,426 *6$ because what you found is that there are $23,426$ ways to form a group of $3$ (combinations of groups of $3$), but within each group of $3$ there are $6$ different ways to order those three people in first, second, and third place (call the people a,b,c and see how many ways you can arrange those 3 letters) – majmun Aug 29 '16 at 03:21

1 Answers1

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The order literally matters: 1st, 2nd, and 3rd place. We have 53 choices for first, then 52 for second once we've chosen 1st, and finally 51 for 3rd. Hence, we have $53\cdot 52\cdot 51 = 140556$ ways that there can be a first, second and third place.

Taking a "combination" of three drivers just counts how many groups of three you can make. It does not take into account the fact that there must be a first, second and third. If the question actually says "combination of three", then sure ${_{53}\mathsf C}_3 = 23426$.

Em.
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  • Certainly if I am a racer I care whether I won or placed third. There is a big difference in the glory awarded. – Ross Millikan Aug 29 '16 at 03:25
  • Ohhh, did I misinterpret the problem? It's very possible. Are you saying it's asking for how many groups of three? – Em. Aug 29 '16 at 03:28
  • No, I think you understood it correctly. I think OP was seduced by the word "combinations" to think order did not matter. I was supporting your answer and gave you an upvote. – Ross Millikan Aug 29 '16 at 03:29
  • Yes, turns out I needed to use permutation instead of combination. Thanks guys, my bad. – buzzard Aug 29 '16 at 03:32
  • @RossMillikan Thank you. But I am curious. This is how I understood it, but now that I've thought about it again, isn't it possible to interpret it as counting the number ways we can make a "winners' circle" (just picking 3 people to be 1st, 2nd, and 3rd)? How else would we ask to count the number of ways to finish 1st, 2nd, 3rd? "How many ways to finish first, second, third?" for example? – Em. Aug 29 '16 at 03:33
  • May be worth pointing out that an alternative way to get the answer was to first grab the set of people who are going to be top 3 and then permute them, so you get (53 choose 3)*3! for a total of 140556. – JKEG Aug 29 '16 at 03:34
  • @user1971507 Are you sure? Do you have the answer/solution? – Em. Aug 29 '16 at 03:34
  • Yes the correct answer is 140,566, since order does matter. If the order didn't matter it would be 23,426. – buzzard Aug 29 '16 at 03:38
  • @user1971507 Well, that's how I understood it anyway. But now I feel like it is ambiguous (despite knowing the "right" answer), since this could be one way of saying "Count the number of winners' circles possible" like I commented above. – Em. Aug 29 '16 at 03:41
  • If you read "combinations" pedantically, implying that order doesn't matter, the answer ${53 \choose 3}=23426$ is correct. With the reference to NASCAR I think order matters. This is now an English question, not a math one. As with many English questions, the answer is not as absolute as with proper math ones. I have lodged my vote, which agrees with your original answer. Others may disagree. – Ross Millikan Aug 29 '16 at 03:43
  • @RossMillikan Yes, I see your point. Thank you. – Em. Aug 29 '16 at 03:45