Find the sum of all numbers greater than $10,000$ formed by using all digits $0,1,2,4,5$ and no digit being repeated in any number.

I could find that the number of such numbers are $96$ but I could not find their sum.



You are correct that there are $96=4!\cdot4$ numbers as they can't start with $0$.

Now write down all the combinations and add them.

As an example, for $1$, we have $$\begin{align}\color{red}{10245\\10254\\10425\\10452\\10524\\10542}\\\color{blue}{12045\\12054\\12405\\12450\\12504\\12540}\\\color{green}{14025\\14052\\14205\\14250\\14502\\14520}\\\color{purple}{15024\\15042\\15204\\15240\\15402\\\hspace{-0.5cm}\color{black}+15420}\\\hline313326\end{align}$$


Hint: imagine writing all of them up, one below the other, as though you're about to add them up the way you (probably) learned in elementary school. How many have a 1 in the first column? A 2? A 3? What does that column sum to? Do this for each column, then put together the results.


Let's keep placeholders for where we will keep the digits :

[] [] [] [] []

Now, in the first box i.e: ten thousands place, we can put any number in the following set $ \{1,2,4,5 \}$ and after that have 4! ways of filling in the remaining numbers. So, there are 4! numbers with 1 as first number, 4! numbers with 2 as first number.. till 5. So, the sum of digits in ten thousand place is given as:

$$ 4!(1+2+4+5) \cdot 10^4$$

Similarly, suppose we fix some digit in the thousand place, then we have $ 3 \cdot 3!$ way of arranging the rest of numbers (remember we can't keep zero in first place). Hence the sum of no.s in thousand place is given as:

$$ 3 \cdot 3! ( 1 + 2 + 4 +5) \cdot 10^3$$

And, similar results for hundreds , tens and ones place. Now, to get the whole number, we just need to add up the parts of it's expanded form. Hence,

$$ \text{number} = 4!(1+2+4+5) \cdot 10^4 + 3 \cdot 3! (1+2+4+5) \cdot 10^3 + 3 \cdot 3! (1+2+4+5) \cdot 10^2 + 3 \cdot 3! (1+2+4+5) \cdot 10^1 + 3 \cdot 3! (1+2+4+5) \cdot 10^0 = 3119976$$


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