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Find sum of 5-digit numbers that can be formed using 0, 0, 1, 2, 3, 4.

I tried solving it by cases but I don't understand how to deal with Identical digits.

Hints would be more appreciated as I want to solve this problem on my own.

Question source: Pathfinder for Olympiad Mathematics by Vikash Tiwari

The Social Pi
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    Hınt : how many number of length $5$ are there starting with $1,2,...$.how many number of length $5$ are there the second digit is $0,1,2,...$ etc – Not a Salmon Fish May 20 '22 at 14:17
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    Try asking the following questions: how many five digit numbers can you form? To answer that, consider the question of how many six digit numbers can you form and seeing what the difference between that problem is and the problem of counting how many five digit numbers there are. Recall that to be called a five-digit number, the leading digit must not be a zero. What proportion of the numbers have $1$ as a leading digit? $2$ as a leading digit? etc... What proportion of numbers have $1$ as the second digit? $2$ as the second digit? etc... – JMoravitz May 20 '22 at 14:17
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    BTW , please add your effort into the question – Not a Salmon Fish May 20 '22 at 14:17
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    Is zero allowed to be the first digit? – Thomas Andrews May 20 '22 at 14:35
  • Use https://math.stackexchange.com/questions/1819629/find-the-sum-of-all-4-digit-numbers-formed-by-using-digits-0-2-3-5-possib?rq=1 as a reference. – bittahProfessional May 20 '22 at 14:47
  • In fact, This SE answer will directly answer your question (only a very slight change is necessary). Since you don't want the answer, just a hint, here is my hint: Most numbers can be formed simply by all permutations of the digits, then you have to account for some numbers that aren't unique because of the two zeroes, and some numbers that don't really count "because of something about the nature of the number zero." There, that should help! – bittahProfessional May 20 '22 at 14:55
  • Hint: Consider the average value of each digit. If the first digit is allowed to be zero, the average value for each digit is the same. If the first digit is not allowed to be zero, you have to figure out that average separately, but the other averages are the same. Let $a$ be the average of the other digits, and $b$ be the average of the first digit. Then the sum is $(10000b+1111a)\cdot N,$ where $N$ is the total number of such $5$-digit numbers. – Thomas Andrews May 20 '22 at 15:08

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