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Factorial Question

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by awilhelm » Sun Dec 14, 2008 8:26 pm
Which of the following is an integer?

I. 12!/6!
II. 12!/8!
III. 12!/(7!)(5!)

Is there a quick way to calculate whether III is an integer? After canceling out 7-1 I'm left with

(12*11*10*9*8)/(5*4*3*2*1)

Thank you!
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by acecoolan » Sun Dec 14, 2008 8:42 pm
Cancel from numerator and denominator

(12*11*10*9*8)/(5*4*3*2*1)

so 5*2 is 10 - that cancels
and 4*3 is 12 - that cancels ..u r left with 3 integers in the numerator. Hence III. 12!/(7!)(5!) is also an integer
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by awilhelm » Sun Dec 14, 2008 8:51 pm
Thanks!
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by ronniecoleman » Wed Dec 17, 2008 11:14 pm
12! / 5! 7!

Have a rule if a!/b!d! if a = b+ d

then a!/b!d! is a integer !
otherwise...we need to check it out
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by parallel_chase » Thu Dec 18, 2008 3:29 am
Here is another rule

Product of n consecutive integers will always be divisible by n!

12! /5! 7!

we know that 7! cancels out, whats left is

(8*9*10*11*12)/5!

if we follow the above rule, we have product of 5 consecutive integers, therefore the result will be an integer.

Hope this helps.
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Re: Factorial Question

by PVtrainer » Thu Dec 18, 2008 9:20 am
awilhelm wrote:Which of the following is an integer?

I. 12!/6!
II. 12!/8!
III. 12!/(7!)(5!)

Is there a quick way to calculate whether III is an integer? After canceling out 7-1 I'm left with

(12*11*10*9*8)/(5*4*3*2*1)

Thank you!
The value of option III is nothing but the value of 12C7 which always comes as integer or you can see that after canceling 7! we are left with 5! in denominator which is equal to 120 and in numerator it comes from product of 12 and 10 so it comes as integer . Hence all the values are integers
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