Q13/118) How many ordered quadruples of positive integers(w,x,y,z) are there such that w!=x!+y!+z!?

w cannot be 4 or more

as x = 3, y = 3, z =3 gives x! + y! + z! = 18(maximum) < 4 !

w = 3 ! =< x! + y!+z! = 6( x =2 , y = 2, z= 2) no more solution

w = 2! = 2 and so no solution as x = 1, z = 1, y = 1 give 3

w cannot be 1 then x,y,z cannot take any value

so one solution w = 3, x=y=z = 2