For all M's and N's - 23/04/2013

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For all M's and N's - 23/04/2013

Postby showmyiq » Tue Apr 23, 2013 8:11 am

Can you prove that

(m+n)! is dividable by m!n!

for all integer numbers m and n?

m! defines m factorial
m! = 1*2*3*...*m
examples:
1!=1
2!=1*2
3!=1*2*3
etc.
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Re: For all M's and N's - 23/04/2013

Postby aam » Tue Apr 23, 2013 1:26 pm

Isn't it (m+n)! divisible by m!n!?
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Re: For all M's and N's - 23/04/2013

Postby showmyiq » Tue Apr 23, 2013 1:57 pm

Thanks for your correction - I have replaced their position by mistake.
I will inverse them right now.
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Re: For all M's and N's - 23/04/2013

Postby showmyiq » Tue Apr 23, 2013 2:00 pm

Just another task – if m!n! is dividable by (m+n)!, what conclusions we can make for m and n?
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Re: For all M's and N's - 23/04/2013

Postby aam » Tue Apr 23, 2013 5:06 pm

We can say:
(m+n)!=(m+n)(m+n-1)(m+n-2)...(m+n-(n-1)) x m!
clearly (m+n)! is dividable by m!, and we got:
(m+n)!/m!=(m+n)(m+n-1)(m+n-2)...(m+n-(n-1)).
Now, the right hand side of the last equation consists of multiplication of n consecutive integers. Among them, there exist integers dividable by n, n-1, n-2, ... , 1. So, the right hand side is dividable by n(n-1)(n-2)...1 = n!,
And we can conclude (m+n)! is dividable by m!n!.
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Re: For all M's and N's - 23/04/2013

Postby showmyiq » Tue Apr 23, 2013 5:34 pm

Short and elegant solution!
Your name will be published as the one who solved the problem first.

You can share some puzzles or math problems in the forum too. I will be happy to publish them in the main site.
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Re: For all M's and N's - 23/04/2013

Postby showmyiq » Wed Apr 24, 2013 7:21 am

The problem was archived here http://www.showmyiq.com/2013/April/23-04-2013-For%20all%20Ms%20and%20Ns.html

I just want to add another possible solution.
If we analyze the Newton Binomial Coefficients, we will see that all are represented in the same format in our problem. Since all the Binomial Coefficients are integer numbers, we can conclude that (m+n)! divided by m!n! will be always a natural number.

For reference you can see them here: http://en.wikipedia.org/wiki/Binomial_coefficient
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