Factorising

Oct 2012
751
212
Ireland
They use the fact that 5! = 5* 4! and 4!= 4*3! to change 8!/(4!*4!) into 8!/(4!*(4*3!)) and 8!/(5!*3!) into 8!/((5*4!)*3!) to get 4!3! in the denominator of both quantities then factorise 8!/(4!3!)
 
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Jan 2013
30
6
India
for multiplicatin instead of (×), ( . ) is used
8!/4!.4! + 8!/5!.3!
= 8!/4!.4! + 8!/4!.5.3!
= 8!/4![1/4! + 1/5.3!]
= 8!/4![1/3!.4 + 1/5.3!]
= 8!/4!.3![1/4 + 1/5]
----
 
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Prove It

MHF Helper
Aug 2008
12,897
5,001
for multiplicatin instead of (×), ( . ) is used
8!/4!.4! + 8!/5!.3!
= 8!/4!.4! + 8!/4!.5.3!
= 8!/4![1/4! + 1/5.3!]
= 8!/4![1/3!.4 + 1/5.3!]
= 8!/4!.3![1/4 + 1/5]
----
Actually, a (.) is used as a decimal point, it's actually a centred dot, \(\displaystyle \displaystyle \begin{align*} \cdot \end{align*} \) which is used for multiplication. Though a \(\displaystyle \displaystyle \begin{align*} \times \end{align*} \) is also acceptable as long as it doesn't get mixed up with any \(\displaystyle \displaystyle \begin{align*} x \end{align*} \) variables.
 
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topsquark

Forum Staff
Jan 2006
11,575
3,454
Wellsville, NY
does anybody know how this came from?

thank you
The solution has already been given, but I thought I'd rewrite it a bit:
\(\displaystyle \frac{8!}{4! \cdot 4!} + \frac{8!}{5! \cdot 3!} = \frac{8!}{4! \cdot 3!} \left ( \frac{1}{1 \cdot 4} + \frac{1}{5 \cdot 1} \right )\)

We have a forum that deals with how to write expressions. It's the LaTeX Help forum. It's pretty easy to learn the basics. :)

-Dan
 
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