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rewrite the following using fatorial notation (show working) (n+5)(n+4)(n+3)(n+2)(n+1)  

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barcelonamessi | Student, Grade 11 | (Level 1) eNoter

Posted February 9, 2013 at 9:52 AM via web

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rewrite the following using fatorial notation (show working)

(n+5)(n+4)(n+3)(n+2)(n+1)

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 9, 2013 at 12:20 PM (Answer #1)

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You need to multiply and divide `(n+5)(n+4)(n+3)(n+2)(n+1)` by the product `n*(n-1)*(n-2)*...*3*2*1` such that:

`(n+5)(n+4)(n+3)(n+2)(n+1) = ((n+5)(n+4)(n+3)(n+2)(n+1)*n*(n-1)*(n-2)*...*3*2*1)/(n*(n-1)*(n-2)*...*3*2*1)`

You need to remember that `n*(n-1)*(n-2)*...*3*2*1 = n!` and `1*2*3*...*n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5) = (n+5)!`

`(n+5)(n+4)(n+3)(n+2)(n+1) = ((n+5)!)/(n!)`

Hence, you may write `(n+5)(n+4)(n+3)(n+2)(n+1)` using factorial notation, such that `(n+5)(n+4)(n+3)(n+2)(n+1) = ((n+5)!)/(n!).`

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ahalperi | Middle School Teacher | (Level 1) Adjunct Educator

Posted February 16, 2013 at 9:26 AM (Answer #2)

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  Because  (n+5)! = (n+5)(n+4)(n+3)(n+2)(n+1)n!  (by definition)

 

If we choose to multiply by one in the form of   n!/n! then

 

 (n+5)(n+4)(n+3)(n+2)(n+1)     =   

     

                                                    n!                  

  (n+5)(n+4)(n+3)(n+2)(n+1) •    _____    =

                                                    n!

 

     

  (n+5)(n+4)(n+3)(n+2)(n+1)n!           (n+5)!     
   _________________________  =   _______

                  n!                                       n!





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