Ans is D
Power of 10 in an expression gives the number of zeros at the end
To get zeros at the end we should look for 10's because in any expression presence of 10 contributes to a 0. eg:- 6x5x2x3 this expression when solved will have one zero at the end because one 10 is present (5x2= 10)
Observe that looking for 10's means looking for 2's and 5's.
Now let's evaluate given expression
2! = 2x1 here we can't get 10 because there is no 5
4!= 4x3x2x1 here also we can't get 10 because there is no 5
6!= 6x5x4x3x2x1 here we have one 5 and one 2, giving us one 10. Therefore $(6!)^{6!}$ gives $(10)^{6!}$ i.e. 6! zeros
8! =8x7x6x5x4x3x2x1 also has one 5 and one 2, giving us one 10. Therefore $(8!)^{8!}$ gives $(10)^{8!}$ i.e. 8! zeros
10!=10x9x8x7x6x5x4x3x2x1 gives 2 10's (10x5x2) i.e 100 . Therefore $(10!)^{10!}$ gives $(100)^{10!}$ = $(10)^{2\times10!}$ i.e. 2 x (10)! zeros
thus d) 6! + 8! + 2(10!)
