(n-1)!=5-----(n-3)!

Question

(n-1) is the numerator and (n-3)! is the denominator

 

they equeal 5

 

I need to solve and verify

Answer 1

(n - 1)! = (n -1)(n-2)(n- 3)(n-4)..........1

 

(n-3)! = (n-3)(n-4)(n-5)(n-6)........1

 

 

(n -1)(n-2)(n- 3)(n-4)..........1      = 5

(n-3)(n-4)(n-5)(n-6)........1

 

(n - 1)(n- 2) = 5

 

n² - 3n + 2= 5

 

n² - 3n + 2 - 5 = 0

 

n² - 3n -3 = 0

 

????? n is supposed to be an integer. Are you sure you have the right problem?