Find the sum of all the no.'s less than 10000 that can be formed using digits 1,2,3,4 without repetation ? In this question solution is provided as 3 digit number sum = 2!*(6+7+8+9)*111=6660...but in the question it is given that we have to use only 1,2,3,4 digits...please elaborate the explaination

For 4 digit numbers, we have to use all the numbers. sum => 3!*(1+2+3+4)*1111 = 66660
For 3 digit numbers, we have to use 3 of the 4 numbers. If the three numbers are 1,2 and 3, the sum is 2!*(1+2+3)*111 = 2!*(6)*111
If the three numbers are 1, 2 and 4, the sum is 2!*(1+2+4)*111 = 2!*(7)*111
If the three numbers are 1, 3 and 4, the sum is 2!*(1+3+4)*111 = 2!*(8)*111
If the three numbers are 2, 3 and 4, the sum is 2!*(2+3+4)*111 = 2!*(9)*111
The total sum is therefore = 2!*(6+7+8+9)*111 = 6660
Similarly, for 2 digit numbers, the sum => 1!*(3+4+5+5+6+7)*11 = 330
and for 1 digit numbers, the sum is 1+2+3+4 = 10
=> Total sum = 66660 + 6660 + 330 + 10 = 73660