SOLUTION: Find the sum of all positive six-digit integers whose digits are a permutation of the digits in the number 123456 without listing them all.

Algebra ->  Permutations -> SOLUTION: Find the sum of all positive six-digit integers whose digits are a permutation of the digits in the number 123456 without listing them all.       Log On


   

Question 1078654: Find the sum of all positive six-digit integers whose digits are a permutation of the digits in the number 123456 without listing them all.
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
Answer. 111111*21*120.

Solution 

In all, there are 6! = 1*2*3*4*5*6 = 720 such 6-digit numbers.


If you will write all these 720 6-digit numbers in one VERY long column, then

    1.  you will have 120 numbers that have "1" in the "ones" position;
              120 numbers that have "2" in the "ones" position;
              120 numbers that have "3" in the "ones" position;
              120 numbers that have "4" in the "ones" position;
              120 numbers that have "5" in the "ones" position;
              120 numbers that have "6" in the "ones" position.

        If you do sum all these 720 numbers, you will get the contribution to the sum from the "ones" position  
        of the value (1+2+3+4+5+6) = 21 repeated 120 times.


    2.  you will have 120 numbers that have "1" in the "tens" position;
              120 numbers that have "2" in the "tens" position;
              120 numbers that have "3" in the "tens" position;
              120 numbers that have "4" in the "tens" position;
              120 numbers that have "5" in the "tens" position;
              120 numbers that have "6" in the "tens" position.

        If you do sum all these 720 numbers, you will get the contribution to the sum  from the "tens" position 
        of the value (1+2+3+4+5+6)*10 = 21*10 repeated 120 times.


    3.  you will have 120 numbers that have "1" in the "hundreds" position;
              120 numbers that have "2" in the "hundreds" position;
              120 numbers that have "3" in the "hundreds" position;
              120 numbers that have "4" in the "hundreds" position;
              120 numbers that have "5" in the "hundreds" position;
              120 numbers that have "6" in the "hundreds" position.

        If you do sum all these 720 numbers, you will get the contribution to the sum from the "hundreds" position  
        of the value (1+2+3+4+5+6)*100 = 21*100 repeated 120 times.

    4.  Similar for "thousands" position;


    5.  Similar for the "ten thousands" position;


    6.  Similar for the "hundred thousands" position.


Finally, you will have in the total sum 
         the contribution of 21*120          from the "ones" position; 
         the contribution of 21*120*10       from the "tens" position; 
         the contribution of 21*120*100      from the "hundreds" position; 
         the contribution of 21*120*1000     from the "thousands" position; 
         the contribution of 21*120*10000    from the "tens thousands" position; 
         the contribution of 21*120*100000   from the "hundred thousands" position. 


Now sum 6 lines above, and you will get the final answer, which I wrote at the very beginning.

Solved.