2n!=n!n+2! find n pls help

Question

barrycarter · Answer

So, 2n*(2n-1)*(2n-2)*... = n*(n-1)*(n-2)*...*(n+2)*(n+1)*n*(n-1)... ?

anonymous · Answer

yes

anonymous · Answer

its about combinations mostly .. not premutations

barrycarter · Answer

I suspect it's a very small number. Try 0,1,2,3...

barrycarter · Answer

It's (2n)! and not 2*(n!) right?

mathmate · Answer

There is a solution for (2n)!=n!(n+2)!
Try a solution by trial and error.
When they are not equal, see if the inequality changes direction.
If it does, there is probably a solution between the two tries.
As @barrycarter said, the solution is a relatively small number.

anonymous · Answer

If (2n)!=nn!+2, then

n=1.26676271264525267756372262483004386343358269981342...

mathmate · Answer

There is an integer solution if the question was:
(2n)!=n!(n+2)!

anonymous · Answer

It's all about parentheses.

mathmate · Answer

Yep, parentheses are great time savers for everyone, if the question is parenthesized properly to begin with.

anonymous · Answer

Divide it all by n!

anonymous · Answer

oh i ll try that

anonymous · Answer

I think you get (2n)(2n-1)...(n+3)=1. Not that helpful, but it seemed it might have helped.

anonymous · Answer

$$(2n)!=n\times n!+2$$?

anonymous · Answer

this is impossible

clearly for $n>2$$$(2n)!>n\times n!+2$$so check $n=1,2$ only

anonymous · Answer

Sorry, I thought it was (2n)!=n (n+2)!

mathmate · Answer

Please check the question.  I think it is:
(2n)!=n!(n+2)!
By inspection, n cannot be large.
and the solution is an integer n<5.