Find the sum of all the four digit numbers formed using the

We can form a total of 4! or 24 numbers.When we add all these numbers ,let us look at the contribution of of the digit 2 to the sum.

When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed
with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000

Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200

Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20

Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10

The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222
In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555

Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111


Now we can generalize the above

If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to
(n-1)! * ( sum of the n digits ) *( 1111...n times)


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