If n is a positive integer such that
n!/(n-2)! = 342, find n.

• (n)(n-1)(n-2)!/(n-2)!=342
  n^2-n-342=0
  (n-19)(n+18)=0
  n=19,-18
• 12 years agoHelpfull: Yes(4) No(1)
• n = 19
  
  We write n! = n*(n-1)(n-2!)
  Therefore
  n!/(n-2)! = n(n-1)*(n-2)!/(n-2)! = n(n-1).
  
  --> n(n-1) = 342
  --> n^2 -n - 342 = 0
  --> n^2- 19n +18n -342 = 0
  --> n(n-19) +18(n -19) = 0
  --> (n-19)(n+18) = 0
  
  Therefore n -19 = 0; n +18 = 0;
  (i.e) n =19 ; n = -18
  
  We want positive integer. So, n=19.
  
  19!/(19-2)! = 342
  19!/17! = 342.
• 12 years agoHelpfull: Yes(2) No(1)
• n = 19
  
  We write n! = n*(n-1)(n-2!)
  Therefore
  n!/(n-2)! = n(n-1)*(n-2)!/(n-2)! = n(n-1).
  
  --> n(n-1) = 342
  --> n^2 -n - 342 = 0
  --> n^2- 19n +18n -342 = 0
  --> n(n-19) +18(n -19) = 0
  --> (n-19)(n+18) = 0
  
  Therefore n -19 = 0; n +18 = 0;
  (i.e) n =19 ; n =18
  
  We want positive integer. So, n=19.
  
  19!/(19-2)! = 342
  19!/17! = 342.
• 12 years agoHelpfull: Yes(1) No(1)
• ans: 19
  
  n(n-1)(n-2)! / (n-2)!
  n(n-1)=342
  n^2 - n -342=0
  n=-18, 19
• 12 years agoHelpfull: Yes(0) No(1)
• (n*(n-1)*(n-2)!)/(n-2)!=342
  n^2-n-342=0
  n^2+19n-18n-342=0
  n=19 (a positive integer)!
• 12 years agoHelpfull: Yes(0) No(1)