Question 472700: find the sum of all 5 digit numbers can be formed with digits 0,1,2,3,4,repated will not allowed.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Assuming that numbers with leading zeros do not count, there are 4*4*3*2*1 = 96 possible numbers. 24 of these numbers begin with a 1, 24 begin with a 2, ..., and 24 begin with a 4.
Consider all the numbers beginning with a 1. The remaining digits are {0,2,3,4}. However, they can appear in any order, so it can make sense to assume that each digit will appear in each placeholder an equal number of times. There are 24 numbers total, so each number will appear in a placeholder six times. The sum of these numbers is (10000)(24) + 6(4+3+2+0)(1000+100+10+1)
Likewise, we do the same with the other leading numbers 2,3,4 to get
(10000)(24) + 6(4+3+2+0)(1000+100+10+1)
(20000)(24) + 6(4+3+1+0)(1000+100+10+1)
(30000)(24) + 6(4+2+1+0)(1000+100+10+1)
(40000)(24) + 6(3+2+1+0)(1000+100+10+1) (see how to obtain these sums?)
Adding them all up, we get
(100000)(24) + 6(30)(1000+100+10+1) = 2599980.
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