<s>[INST] Find the SUM of all the numbers of S. Let S be the set of all three digits numbers. Such that


*

*The digits in each number are all from the set
$\left\{1,2,3,\ldots,9\right\}$.

*Exactly one digit in each number is even.


Then, find the SUM of all the numbers of S.
I have tried that, 
The only even digit can have any of the three positions, 
And the digit itself has $4$ choices $\left(2, 4, 6\ \mbox{or}\ 8\right)$. The other two digits can be filled in $5 \times 5 = 25$ ways.
Then the number of elements  in S is $300$.
But I failed to find the total sum of the elements.  Please help me to find this. Thanks in advance.
 [/INST] Note that if you permute $3$ distinct digits $x,y,z$, then the sum of the six 3-digit number is $222\cdot(x+y+z)$. If you permute $3$ digits $x,x,z$, with $x\not=z$ then  the sum of the three 3-digit number is $111\cdot(2x+z)$. Therefore your sum is
$$222\sum_{x<y\ x,y\in\{1,3,5,7,9\}}\sum_{z\in\{2,4,6,8\}}(x+y+z)+111\sum_{x\in\{1,3,6,7,9\}}\sum_{z\in\{2,4,6,8\}}\cdot(2x+z)$$
that is
$$222\cdot 4\sum_{x<y\ x,y\in\{1,3,5,7,9\}}(x+y)+111\cdot 4\cdot 2\sum_{x\in\{1,3,6,7,9\}}x+(222\cdot 10+111\cdot5)\sum_{z\in\{2,4,6,8\}}z$$
and we finally find
$$888\cdot 100+888\cdot 25 
+2775\cdot 20=166500.$$
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