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How about this approach?

With 4 placed in units place, there are 3! permutations of 123 in thousandths, hundredths and tenths place. That means 4 occurs in units place 3! times.

Similarly 3 2 and 1 each occur in units place 3! times. The sum of all numbers in units place is therefore 3!(4 + 3 + 2 + 1) = 6 * 10 = 60.

The same value happens to occur in tenths, hundredths and thousandths place as well, except that it gets multiplied by a power of 10 according to its position.

So, the total should be 60 + 600 + 6000 + 60000 = 66660.

A simple solution--

3!(4+3+2+1)(1+10+100+1000)..easy logic

Here we don't have to do actual calculation, else we will never get answer in written test. Think of this as mathematical + logical question. Lets say we have all permutation of 1234. It will be 4! = 24. Now if we look at each column of this list carefully, every number will appear for 6 time.
1234
1324
2134
2314
3124
3214
...

Now if you just concentrate on 4th column, "4" is coming 6 times, same way if we list all number each number will come 6 time in 4th column. So sum of 4th column is (4*6 + 3*6 + 2*6 + 1*6) = 60.

Same way, for all other columns sum would be 60.

Now we are just one step far away from answer. And that is consider "carry". 4th column will give 6 as carry to 3rd columns, again in turn 3rd column will give 6 as carry to 2nd column and so on.

Therefore, final answer would be 66660.

Total no. of permutations = 4! = 24
Each digit 1,2,3 and 4 will occur exactly 6 times at each position.
Sum of digits at each position = (1+2+3+4)*6 = 60
Total sum
60000
6000
600
60
---------
66660 = Answer
--------

Consider the unit's digit place,each of 4,3,2,1 appears 6 times each.So at the digit place ,this adds to (4+3+2+1)*6=60.Now 6 gets carried forward to the tens place.Similarly at ten's place,it adds to (4+3+2+1)*6+6(the carry)=66.Now again the 6 is carried forward till we get 66660.

Let assume we have n digits. Each of them will appears on i-th position n!/n times. So the general formula is :
n!/n = 6
(1 + 2 + 3 + 4) * 6 * 4 = 240

mind blowing...

its 4! i.e 4*3*2*1=24
now if we pair the number like 1234+4321=5555
2134+3421=5555
in every case we come across 5555
therefore there will be 12 pairs
hence 12*5555=66660

Here is the java program which will print the combinations and sum of all combinations. Little more info apart from sum.

package com.bala.logical;

import java.util.ArrayList;
import java.util.List;

public class SumCombination {
static String finalString = "";

public static void main(String[] args) {
String s = "1234";
List<String> numList = new ArrayList<String>();
System.out.println(arrangeNumbers(s, numList));
System.out.println("Size of the elements "+numList.size());
int sum = 0;
for(String combination: numList){
sum = sum + Integer.parseInt(combination);
}
System.out.println(sum);
}

public static String[] swap(String s) {
String[] combinations = new String[s.length()];
if (s.length() == 1) {
combinations[0] = s;
return combinations;
}
for (int i = 0; i < s.length(); i++) {
combinations[i] = s.substring(1, s.length()) + s.substring(0, 1);
s = s.substring(1, s.length()) + s.substring(0, 1);
}
return combinations;
}

public static List<String> arrangeNumbers(String s, List<String> numberList) {
String[] combinations = swap(s);
for (int i = 0; i < combinations.length; i++) {
finalString = finalString + combinations[i].substring(0, 1);
if (combinations.length == 1) {
numberList.add(finalString);
} else {
String subS = combinations[i].substring(1,
combinations[i].length());
arrangeNumbers(subS, numberList);
}
finalString = finalString.substring(0, finalString.length() - 1);
}
return numberList;
}
}

it can be done by finding all permutation of the given digits without rep, then add each numbers formed. there are totally 4! numbers to add.

Let number = 1234
let sum = 1234.

1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.
2. Find the largest index l such that a[k] < a[l]. Since k + 1 is such an index, l is well defined and satisfies k < l.
3. Swap a[k] with a[l].
4. Reverse the sequence from a[k + 1] up to and including the final element a[n].
5. sum = sum + newly formed number on step 4.

A simple solution--

3!(4+3+2+1)(1+10+100+1000)..easy logic

Here is the java program which will print the combinations and sum of all combinations. Little more info apart from sum.

package com.bala.logical;

import java.util.ArrayList;
import java.util.List;

public class SumCombination {
	static String finalString = "";

	public static void main(String[] args) {
		String s = "1234";
		List<String> numList = new ArrayList<String>();		
		System.out.println(arrangeNumbers(s, numList));
		System.out.println("Size of the elements "+numList.size());
		int sum = 0;
		for(String combination: numList){
			sum = sum + Integer.parseInt(combination);
		}
		System.out.println(sum);
	}

	public static String[] swap(String s) {
		String[] combinations = new String[s.length()];
		if (s.length() == 1) {
			combinations[0] = s;
			return combinations;
		}
		for (int i = 0; i < s.length(); i++) {
			combinations[i] = s.substring(1, s.length()) + s.substring(0, 1);
			s = s.substring(1, s.length()) + s.substring(0, 1);
		}
		return combinations;
	}

	public static List<String> arrangeNumbers(String s, List<String> numberList) {
		String[] combinations = swap(s);
		for (int i = 0; i < combinations.length; i++) {
			finalString = finalString + combinations[i].substring(0, 1);
			if (combinations.length == 1) {
				numberList.add(finalString);
			} else {
				String subS = combinations[i].substring(1,
						combinations[i].length());
				arrangeNumbers(subS, numberList);
			}
			finalString = finalString.substring(0, finalString.length() - 1);
		}
		return numberList;
	}
}