Question 1021015: Find the sum of all five-digit that can be formed by using the digits 2,3,5,7 and 8,where each of these digits appears in each number.
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! There are 5! = 120 different five-digit numbers that can be formed from the numbers 2, 3, 5, 7, 8.
Note that in each place (in the ten-thousands place, for example), a digit appears exactly 4!=24 times.
The sum of all numbers in the ten-thousands place only is (2*24+3*24+5*24+7*24+8*24)*10,000 = (2+3+5+7+8)24*10,000 = 6,000,000.
The sum of all numbers in the thousands place only is (2*24+3*24+5*24+7*24+8*24)*1,000 = (2+3+5+7+8)24*1,000 = 600,000.
The sum of all numbers in the hundreds place only is (2*24+3*24+5*24+7*24+8*24)*100 = (2+3+5+7+8)24*100 = 60,000.
The sum of all numbers in the tens place only is
(2*24+3*24+5*24+7*24+8*24)*10 = (2+3+5+7+8)24*10 = 6,000.
The sum of all numbers in the ones place only is
(2*24+3*24+5*24+7*24+8*24)*1 = (2+3+5+7+8)24*1 = 600.
Therefore the sum of all the numbers is the sum all the numbers determined above, which is 6,666,600.
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