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i'm currently trying to figure out how i can reduce the size of the numbers below, in such case that i can reason with them. E.g. in this example i would like to figure out the median.

  • A) 1000!
  • B) 500!500!
  • C) 600!400!
  • D) 700!300!
  • E) 800!200!

What I've tried, however i'm not sure this is correct:

1) divide each number by 100

  • 5!*5! = 14400
  • 6!*4! = 17280
  • 7!*3! = 30240 (median)
  • 8!*2! = 80640
  • 10! = 3628800

However it would take some time to calculate 10! for example by hand.

2) figure out the relationship between two numbers by dividing them:

500!500! / 600!400! =

500 x 499 x ... x 402 x 401 / 600 x 599 x ... x 502 x 501

As a result: 600!400! > 500!500!.

But this seems to much work to do this for all numbers...

There is probably an easier way to figure this out. Any hints?

Thanks in advance.

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migrated from mathematica.stackexchange.com Feb 9 '15 at 20:06

This question came from our site for users of Wolfram Mathematica.

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    $\begingroup$ I think you should ask this question @ Mathematics.SE $\endgroup$ – Sektor Feb 9 '15 at 19:45
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It is not correct to divide all the numbers by $100$ without justifying it, though by the argument below you got the right result. If you are looking for the median, you don't need to calculate any of the numbers, just compare them. Intuitively, you could think about Pascal's triangle. We know the numbers in a row are at a maximum in the middle. In the $1000$th row, the end numbers are $\frac {1000!}{1000!0!}=1$. When you go in $200$ places, you get to $\frac {1000!}{800!200!}$, which is another of your numbers. To make the Pascal entry large, you want the denominator small, which tells you that $$1000!\gt 800!200! \gt 700!300! \gt 600!400! \gt 500!500!$$ and the median is $700!300!$. Another way to see this is that in going from $1000!$ to $800!200!$ you divide by $200$ numbers from $801$ through $1000$ and multiply by $200$ numbers from $1$ to $200$. Clearly this will reduce the product. At each stage you are removing large factors and adding small factors, so the claimed ordering is correct.

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