sum of numbers formed by permutations I have digits 2,3,4,5. I have been asked to find the sum of all 4 digits the numbers that can be formed using these digits without repetition such that all are included in the number.

Can someone help me solve this quickly?

The answer is 93324.
Fixing 5 in the units digit, you get $3!=6$ numbers. So 5 comes 6 times in the units digit. Same goes for 2,3,4

hence sum of the units digits = $6(2+3+4+5)=84$

Same argument goes for ten's place, hundreds place and so on. hence the sum=

$84 \times 1000+84 \times 100+84 \times 10+84=84 \times 1111=93324$
For the 2 you get:
$2000 \cdot 3! + 200 \cdot 3! + 20 \cdot 3! + 2 \cdot 3!$
The 1st term corresponds to 2 being at position 1,
the 2nd term corresponds to 2 being at position 2 and so on.   

Then you have the same pattern for 3,4,5.   

So you get: $(2222 + 3333 + 4444 + 5555) \cdot 6 = 93324$