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Question 154621: hey, i would greatly appreciate it if you could solve this combinatorics problem for me
(n+2)!/n! = 56: hey, i would greatly appreciate it if you could solve this combinatorics problem for me
(n+2)!/n! = 56
Answer by edjones(2401) About Me  (Show Source):
You can put this solution on YOUR website!
(n+2)!/n! = 56
((n+2)(n+1)*n!)/n!=56
(n+2)(n+1)=56
n^2+3n+2=56
n^2+3n-54=0
(n+9)(n-6)=0
n=-9, n=6
.
Ed
Question 154621: hey, i would greatly appreciate it if you could solve this combinatorics problem for me
(n+2)!/n! = 56: hey, i would greatly appreciate it if you could solve this combinatorics problem for me
(n+2)!/n! = 56
Answer by jim_thompson5910(9376) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given equation.


Expand (n+2)!. Notice how each successive term is going down by 1



Expand n!




Subtract and simplify



Photobucket - Video and Image Hosting Cancel out the common terms


Simplify


(n+2)(n+1)=56 Reduce


n^2+3n+2=56 FOIL the left side


n^2+3n-54=0 Subtract 56 from both sides.


(n+9)(n-6)=0 Factor the left side


n+9=0 or n-6=0 Set each factor equal to zero


n=-9 or n=6 Solve for "n"



So the possible solutions are n=-9 or n=6. However, you cannot evaluate the factorial of a negative number (well not yet anyway). So the only solution is n=6