sum of all the numbers that can be formed using digits 1,2,3,4 without repetition

i did not get the solution given in the topic test. so please give an alternate approach if possible

Hi Aditya,
When 1 is at thousands position, the other 3 numbers can be arranged in 3! ways so there will be 6 such numbers. For each number value of 1 will be 1000. Similarly when other numbers occupy thousands place, there will be 6 numbers and each time value will be 1000.
Similarly when any occupies 100's position, its value will be 100 and son on.
So required answer is 1*1000*6 + 2*1000*6 + 3*1000*6 +4*1000*6 + 1*100*6 +2*100*6 ....... I hope you get the pattern
After taking the terms common it reduces to (1+2+3+4)*1111*6 = 10*6*1111
In general if a,b,c,d...n are the numbers then general formula for sum is (a+b+c+d...+n)*1111...n times *(n-1)!.
Hope that clears your doubt. If not please revert back with your query
Thanks.

thanks akhilesh