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 October 22nd, 2014, 11:12 AM #1 Newbie   Joined: Oct 2014 From: Here Posts: 17 Thanks: 0 Factorial dividing help Hello, Can you explain to me why is the following equal? [(n!)]²/[(n+1)!]² = 1/(n+1)(n+1) Thanks
October 22nd, 2014, 11:19 AM   #2
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Quote:
 Originally Posted by jamesmith134 Hello, Can you explain to me why is the following equal? [(n!)]²/[(n+1)!]² = 1/(n+1)(n+1) Thanks
$\displaystyle \frac{(n!)^2}{(n + 1)!)^2} = \frac{n!~n!}{(n + 1)!~(n + 1)!}$

$\displaystyle = \frac{n!}{(n + 1)!} \cdot \frac{n!}{(n + 1)!}$

Can you finish from here?

-Dan

October 22nd, 2014, 11:59 AM   #3
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Quote:
 Originally Posted by jamesmith134 [(n!)]²/[(n+1)!]² = 1/(n+1)(n+1)
$\displaystyle \color{blue}{[(n!)]^2/[(n+1)!]^2 =\left[\dfrac{n!}{(n+1)!}\right]^2=\left[\dfrac{n!}{n!(n+1)}\right]^2=\left[\dfrac{1}{n+1}\right]^2=\dfrac{1}{(n+1)^2}=\dfrac{1}{(n+1)(n+1)}}$

Last edited by aurel5; October 22nd, 2014 at 12:01 PM.

 October 22nd, 2014, 01:57 PM #4 Newbie   Joined: Oct 2014 From: Here Posts: 17 Thanks: 0 Thanks you guys it's just that I don't know what is allowed to do when there are roots with factorials My question to you, will you actually do the steps you showed me? Or you have a way to do see it and just write the final answer without the steps?
October 22nd, 2014, 03:31 PM   #5
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Quote:
 Originally Posted by jamesmith134 Thanks you guys it's just that I don't know what is allowed to do when there are roots with factorials My question to you, will you actually do the steps you showed me? Or you have a way to do see it and just write the final answer without the steps?
This is the kind of thing that ruins me in timed exams. Even though I know a few tricks and can do them in my head I usually go ahead and write them down anyway as it's too easy to make mistakes. It's more time consuming, but you learn the material better.

-Dan

 October 22nd, 2014, 05:17 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra Unless the steps are blindingly obvious, you should always write them down. As Dan says, it cuts down on silly mistakes. More to the point, it makes it clear for the person reading it. If what you are writing has even the vaguest chance of looking difficult or confusing, you should add a step in the middle. Thanks from topsquark and jamesmith134
October 22nd, 2014, 05:40 PM   #7
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Quote:
 Originally Posted by jamesmith134 Hello, Can you explain to me why is the following equal? [(n!)]²/[(n+1)!]² = 1/(n+1)(n+1)
jamesmith134, they are not equal. The expression on the right-hand side needs grouping symbols,
such as braces, square brackets or parentheses, around the denominator:

[(n!)]²/[(n+1)!]² = 1/{(n+1)(n+1)}

[(n!)]²/[(n+1)!]² = 1/[(n+1)(n+1)]

[(n!)]²/[(n+1)!]² = 1/((n+1)(n+1))

October 22nd, 2014, 07:46 PM   #8
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Quote:
 My question to you, will you actually do the steps you showed me?
Actually Aurel5 did do the steps. If you have difficulty following the algebra, I suggest to try substituting some positive integer for n. I did it with Aurel5's solution below by substituting 5 for n and writing out the factorials in detail.
Please examine and see how the arithmetic is done.

Quote:
 $\displaystyle \color{blue}{[(5!)]^2/[(5+1)!]^2 =\left[\dfrac{5!}{(5+1)!}\right]^2=\left[\dfrac{5×4×3×2×1}{(5+1)(5×4×3×2×1)}\right]^2=\left[\dfrac{1}{5+1}\right]^2=\dfrac{1}{(5+1)^2}=\dfrac{1}{(5+1)(5+1)}}$

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