If unit digit of x! + y! + z! is 9 then find the value of x!.y!.z!

0! = 1
1! = 1
2! = 2
3! = 6
4!= 24
5! = 120
6! = 720
Hence all factorials greater than or equal to 5 will have 0 at their unit place. so they are not of our use in this question.

in both the cases:-
x!.y!.z! = 0!.2!.3! = 1!.2!.3! = 12 Answer