1
$\begingroup$

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

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ ok.thanks. understood $\endgroup$
    – user791682
    Commented Jul 11, 2020 at 14:32

You must log in to answer this question.