Day 9: The Reindeer Quiz
Santa's nine reindeer Dasher, Dancer, Prancer, Vixen, Comet, Cupid, Donner, Blitzen and Rudolph are tired of pulling sleighs all the time so they are going to have a quiz.
One reindeer will be the quiz master and the others will split into four teams of two.
In how many ways can they arrange themselves?
This problem is part of the Advent Calendar 2015 .
The answer is 945.
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3 solutions
Nice question.
Rudolph will be bad at a trivial quiz, because he "will go down in history".
Did you know that originally Donner was Donder?!?!?!? I had to name 8 reindeer (no rudolph) recently for extra credit on a math test, and my math teacher pointed that out to me. Pretty cool! (Though I think we all prefer Donder.)
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Interesting - I never knew that before!
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Also, the 12 days of Christmas, it is not 4 calling birds. (Well, we changed it to that.) It was orginally 4 collie birda, but no one liked that!
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@Colin Carmody – You must be Santa. Ho! Ho! Ho!
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@Anupam Nayak – Umm, ok I guess? (I'm not, but whatever!)
@Colin Carmody – Nice! In fact I have made a problem inspired from this discussion right here . I have credited you and I hope you enjoy it! :)
In response to Michael Ng : Sir you have used multiplication rule but i m not able to understand how u did pls explain again Sir :)
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Of course :) So at first we must choose the first pair, so we choose 2 out of 9 . Now in any case once we have done that we have 7 left, and now we choose the second pair; 2 out of 7 . We continue that ( 2 out of 5 then 2 out of 3 ) until we have all four pairs.
Why do we multiply? It is because for each choice of the first pair, we are left with the same situation of having 7 reindeer. In this 7 there are always the same number of ways to choose the remaining three pairs. Therefore we must multiply the number of ways to get the first pair by the number of ways to get three pairs from 7 .
Then consider the three pairs from 7 . Using the same logic there are always the same number of ways to choose the remaining two pairs, so we must multiply. Continuing this logic we get the above solution.
Hope this helps; if you need any clarification feel free to ask.
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In response to Michael Ng : Thank you very much it helped me a lot :)
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@Chirayu Bhardwaj – No problem, glad to know that it helped! :)
Cool, I did almost the same thing, except setting aside the quizmaster first, there are 4 ! ( 2 8 ) ( 2 6 ) ( 2 4 ) ( 2 2 ) ways to group the other reindeer. There are 9 options for quizmaster so multiply by 9 .
First, we choose the quiz master.
Then, we have one of the remaining 8 to choose their partner.
Then, we have one of the remaining 6 to choose their partner.
Then, we have one of the remaining 4 to choose their partner.
Then, we have one of the remaining 2 to choose their partner.
9 × 7 × 5 × 3 × 1 = 9 4 5
Nice, this seems to be very elegant! Just that so I can understand it more, can you please explain why you are not double counting any possible arrangement and how you got the calculation?
Thanks!
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That's how I got it too. Clearly, there are 9 ways to choose the quiz master. Once he is chosen, you get one of the others to choose their partner, for which there are 7 choices. Then one of the remaining 6 chooses, for which there are 5 choices, and so on as Bk said.
To see how this avoids double counting, just add the rule that, once the quiz master is chosen, it is the reindeer who comes first alphabetically who gets to choose their partner. Once that is done, the reindeer who is first alphabetically in the remaining 6 gets to choose, and so on until the teams are all fixed. It doesn't matter what rule you use, just as long as you remove each team as they are formed.
For what it's worth, this is called the "double factorial", denoted 9!! (in this case). It's a quick way to find the number of ways to choose unique pairs from pool of distinguishable items.
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Nice! Thank you; that makes it a lot clearer. The double factorial fact is interesting!
First, we choose the quiz master, there are 9 ways to do that. Then, we can list the names of the 8 remaining reindeer in 8! ways.
Now, we choose the first two on the list for the first team, the second two for the second team and so on. As the order of the names within a team does not matter, we have to divide by (2!)^4=2^4=16. Neither does the order of the teams matter, therefore we also have to divide by 4!.
For the reasons described above, our answer is:
9×8!/(16×4!)=945
(Alternatively, we can list all 9 names first and choose either the first or the last (as Michael Ng did) for quiz master. In this case, our formula would be 9!/(16*4!)=945)
Ah yes, a very nice solution! Thank you.
We can choose the four teams of two then the one remaining can be the quiz master.
First we choose 2 out of 9 for the first team, then 2 out of the remaining 7 and so on, giving ( 2 9 ) × ( 2 7 ) × ( 2 5 ) × ( 2 3 ) = 2 2 6 8 0 But now we have counted every team 4 ! times so divide by 2 4 to give 9 4 5 .