Question 154621
*[Tex \LARGE \frac{\left(n+2\right)!}{n!}=56] Start with the given equation.



*[Tex \LARGE \frac{\left(n+2\right)\left(n+2-1\right)\left(n+2-2\right)\left(n+2-3\right)\cdots\left(3\right)\left(2\right)\left(1\right)}{n!}=56] Expand {{{(n+2)!}}}. Notice how each successive term is going down by 1




*[Tex \LARGE \frac{\left(n+2\right)\left(n+2-1\right)\left(n+2-2\right)\left(n+2-3\right)\left(n+2-4\right)\left(n+2-5\right)\left(n+2-6\right)\cdots\left(3\right)\left(2\right)\left(1\right)}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\cdots\left(3\right)\left(2\right)\left(1\right)}=56] Expand {{{n!}}}





*[Tex \LARGE \frac{\left(n+2\right)\left(n+1\right)\left(n\right)\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\cdots\left(3\right)\left(2\right)\left(1\right)}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\cdots\left(3\right)\left(2\right)\left(1\right)}=56] Subtract and simplify




<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/combinatorics.png" alt="Photobucket - Video and Image Hosting">  Cancel out the common terms



*[Tex \LARGE \frac{\left(n+2\right)\left(n+1\right)}{1}=56] 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}}}