Let $S$ be the set of all 3 - digits numbers. Such that (i) The digits in each number are all from the set $\{1,2,3, \ldots ., 9\}$ (ii) Exactly one digit in each number is even The sum of all number in $\mathrm{S}$ is
My approach: The sum of the digits in unit place of all the numbers in s will be same as the sum in tens or hundreds place. The only even digit can have any of the three positions, i.e. $^{3} c_{1}$ ways And the digit itself has 4 choices $(2,4,6 \text { or } 8) .$ The other two digits can be filled in $5 \times 4=20$ ways. Then the number of numbers in $\mathrm{S}=240$
Number of numbers containing the even digits in units place $=4 \times 5 \times 4=80$
The other 160 numbers have digits 1,3,5,7 or 9 in unit place, with each digit appearing
$\frac{160}{5}=32$ times. Sum in units place $=32(1+3+5+7+9)+20(2+4+6+8)$
$=32.5^{2}+20 \times 2 \times \frac{4 \times 5}{2}=32 \times 25+20 \times 20=1200$
$\therefore$ The sum of all numbers $=1200\left(1+10+10^{2}\right)=1200 \times 111=133200$
NB: Find the SUM of all the numbers of S. I have found this In SE.But this doesnot match my answer.Am I wrong? Please tell me if any better aprroach is there
You are assuming that the two odd digits must be different. The other question assumes that they can be the same. So your $5\times 4$ is their $5\times 5$.
As far as I can see there is nothing in the question to suggest that those digits must be different. So I prefer the other answer.