Question 594526
*[tex \LARGE \frac{3n(n-2)!}{n-3} = 105], Divide through by 3


*[tex \LARGE \frac{n(n-2)!}{n-3} = 35]


We can write n(n-2)! as n!/(n-1). Above equation is equivalent to


*[tex \LARGE \frac{n!}{(n-1)(n-3)} = 35] or *[tex \LARGE n(n-2)(n-4)! = 35]


Here it is obvious that n should be fairly small (at most 8 because (8-4)! = 24). However, n must be at least 7 (because n(n-2)(n-4)! ≡ 0 (mod 7)). Therefore there are no solutions.