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Easier way of simplifying this factorial equation?

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Easier way of simplifying this factorial equation?

by ssy » Sat Oct 20, 2007 10:36 pm
Is there a way to tell if 12! / (7!5!) is an integer without fully working it out?

In my calculations, I've only managed to simplify it to:

12 x 11 x 10 x 9 x 8 / 5 x 4 x 3 x 2 x 1

..which would still take around 40 precious seconds to work out if it equals an integer..is there a shortcut?
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Re: Easier way of simplifying this factorial equation?

by jimbow » Mon Oct 22, 2007 4:37 am
ssy wrote:Is there a way to tell if 12! / (7!5!) is an integer without fully working it out?

In my calculations, I've only managed to simplify it to:

12 x 11 x 10 x 9 x 8 / 5 x 4 x 3 x 2 x 1

..which would still take around 40 precious seconds to work out if it equals an integer..is there a shortcut?

you've got it:

5x2=10
4x3=12

you simplify and you have 11x9x8 / 1 (you don't need to calculate this). You know that 11x9x8 is an integer (product of integers)
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by ssy » Mon Oct 22, 2007 4:49 am
Great, thanks guys!
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by agni_mba » Mon Oct 22, 2007 3:00 pm
another way of looking at this is: 12!/(7!.5!) is 12C5 i.e. no. of different ways of picking 5items from a set of 12. Obviously that can't be a non-integer...

cheers
agni
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