Let's take a look here ,

(1/x) + (1/y) =(1/n!) number of solutions

let's n! as p okk ? Now ..

(1/x) + (1/y) =(1/p)

(x+y)/(xy) = (1/p)

 px + py =xy

- px -py +xy=0

 p^2 - px -py +xy =p^2   [ adding p^2 both sides ]

 p(p-x) -y(p-x)=p^2

 (p-y) (p-x) =p^2

 Now , think (p-y) as A and (p-x) as B

 So ,   A*B = p^2 

 Now , if anybody asks u the number of solutions of this solution .
 Then It will make U common sense that the number of divisors will be unique solutins Because Look At this . Let's think p^2 =36

 So solutions are (value of A and B) 

 a * b

 1*36=36
 2*18=36
 3*12=36
 4*9=36
 6*6=36 


 a * b

 36*1=36
 18*2=36
 12*3=36
 9*4=36

 Total 9 unique solutions . 

 Now , put n! in p So , U have to find out the number of divisors of (n!)^2 . That's all !!! 

 Hope it helps 

 My github link 

https://github.com/joy-mollick/Problem-Solving-Solutions-Math-Greedy-