Question 241099
{{{(n+2)!/n! = 56}}}
{{{(n+2)! = 56 * n!}}}
{{{(n+2)(n+1)n! = 56 *n!}}}}
{{{(n+2)(n+1) = 56}}}
{{{n^2 + 3n -54 = 0}}}
{{{(n+9)(n-6) = 0}}}
So n = -9 or n=6

Since n=-9 is not possible in n!, then it must be n=6
Which makes sense since n+1=7  and n+2=8 and 7*8 = 56