Trick of the day

Sum of numbers of numbers formed using n digits. 

Sum = Sum of digits * (n-1)! * 1111.... n times.
                    provided no digit is zero & digits are not repeated.  


e.g. Find sum of all digits formed using 2,3,4 & 6 such then digits are not repeated. 

(1)  99990  (2)  62525  (3)  99900  (4)  98900  (5) None of these.

Solution :-
Since zero is not one of digits & digits are not repeated, we can use formula 

Sum = Sum of digits * (n-1)! * 1111.... n times.

sum =  15 * 3!* 1111 = 99990. Hence option 1