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I'm having trouble using a Permutation formula for finding out how many different ways there are to seat 264 people at 481 desks. The trouble I'm having is that in the Permutation formula (nPr = n! / (n-r)!) n would be the distinct people (264) and r would be the number of desks (i.e. spots) to fill (481).

But this calculation doesn't work, as 264 - 481 = (-217), for which you cannot calculate the factorial.

So my question is, is it possible to calculate the number of different ways a number of distinct objects can be placed into a greater number of places, as with my example above? Can r be greater than n in the Permutation formula? And if so, how?

Thanks in advance for any help.

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1 Answer 1

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I think you can just think of it as how many ways can I rearrange 264 people and 217 "empties" Something like $\frac{481!}{217!}$

481! ways to arrange the seats, but 217! ways the empties could be arranged identically.

Another way would be to think of it like this: Person 1 can be seated in any of 481 seats. Person 2 has 480 choices.

$481*480*479*\dots*(481-263)=\frac{481!}{217!}$

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  • $\begingroup$ So if I read this correctly, the calculation could work if I think of the seats as the distinct objects, and the people as the "spots" they are being assigned to? $\endgroup$
    – Night1505
    Dec 5, 2014 at 19:49
  • $\begingroup$ Yes. Imagine place cards being handed out to people. $\endgroup$ Dec 5, 2014 at 19:50
  • $\begingroup$ Well I'll be... I had a tugging suspicion that it could be worked out by just changing my thinking, but for the life of me couldn't work it out. Probably a "Couldn't see the forest for the trees" sort of thing. Thank you so much for the patient and super quick response. :) $\endgroup$
    – Night1505
    Dec 5, 2014 at 19:53
  • $\begingroup$ No problem. It works, as you probably see, because the seats really are different. Sitting in the front row center is different than back row. $\endgroup$ Dec 5, 2014 at 19:55

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