If unit digit of $a!+b!+c!$ is 9 then find the value of ${(a!)(b!)(c!)}$
Choose the correct answer
A
8
B
5040
C
0
D
12
Hint
Use basic concept of factorial value
Step-by-Step Solution
Since, unit digit of all the factorials more than 4 is 0, hence unit digit is given by either $0!+2!+3!$ or $1!+2!+3!$. In both the cases the value of ${(a!)(b!)(c!)}$ is $1\times2\times6=12$