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August 16th, 2014, 07:33 AM  #1 
Newbie Joined: Jun 2014 From: earth Posts: 11 Thanks: 0  Need Help w/ Factorial Simplification
Hello, I am trying to simplify questions such as: (n + 1)!/n! (n + 2)!/n! ((n + 1)! + n!  (n  1)!)/(n  1)! Naturally I don't want answers for them all, however, I do know that the first one equals n + 1, and the second one equals (n + 2)(n + 1), but I don't see how to get to that answer. If anyone can do out the simplification process simply for me, that would be greatly appreciated, thanks! 
August 16th, 2014, 12:58 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716  Quote:
(n+1)! = 1.2.3......n.(n+1) = n!.(n+1)  
August 16th, 2014, 04:28 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, teetar! Do you understand basic factorials? $\quad \dfrac{5!}{4!} \;=\;\dfrac{5\cdot\cancel{4\cdot3\cdot2\cdot1}}{ \cancel{4 \cdot3\cdot2\cdot1}} \;=\;5$ That is: $\:\dfrac{5!}{4!} \;=\;\dfrac{5\cdot\cancel{4!}}{\cancel{4!}} \;=\;5$ Quote:
$\dfrac{(n+1)!}{n!} \;=\;\dfrac{(n+1)\,\cancel{n!}}{\cancel{n!}} \;=\;n+1$ Quote:
$\dfrac{(n+2)!}{n!} \;=\;\dfrac{(n+2)(n+1)\,\cancel{n!}}{\cancel{n!}} \;=\;(n+2)(n+1)$ Quote:
$\dfrac{(n+1)! + n!  (n1)!}{(n1)!} \;\;=\;\;\dfrac{(n+1)!}{(n1)!} + \dfrac{n!}{(n1)!}  \dfrac{(n1)!}{(n1)!} $ $\qquad =\;\;\dfrac{(n+1)(n)\cancel{(n1)!}}{\cancel{(n1)!}} + \dfrac{n\cancel{(n1)!}}{\cancel{(n1)!}}  \dfrac{\cancel{(n1)!}}{\cancel{(n1)!}} $ $\qquad =\;\; (n+1)n + n  1 \;\;=\;\;n^2 + 2n  1$  
August 18th, 2014, 11:27 AM  #4 
Newbie Joined: Jun 2014 From: earth Posts: 11 Thanks: 0 
Thanks a bunch for your help!


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